Work and Energy

Question 1

Question 1: 1. As shown in the figure is a trampoline athlete performing in the air, in the process of the athle...

1. As shown in the figure is a trampoline athlete performing in the air, in the process of the athlete from the lowest point of the beginning of the rebound to the imminent separation from the trampoline, the following statements are correct ![](/images/questions/phys-work-energy/image-001.jpg)

A. A. Trampolines have increased bouncing force on people
B. B. Increase in elastic potential energy
C. C. increase in kinetic energy
D. D. Increase in gravitational potential energy

Answer

Answer: D

Solution: In the process of the athlete's rebound from the lowest point to the process of separation from the trampoline, the elastic force decreases, so A is wrong; in the process of the athlete's rebound from the lowest point to the process of separation from the trampoline, the amount of deformation decreases, the potential energy of elasticity decreases, so B is wrong; the combined force is upward first, and then downward, and the velocity increases first and then decreases, the kinetic energy increases first and then decreases, so C is wrong; in the process of the athlete's rebound from the lowest point to the process of separation from the trampoline, the athlete has been moving upward, and the gravitational potential energy increases. In the process of the athlete's rebound from the lowest point to the separation of the trampoline, the athlete has been moving upward, the gravitational potential energy increases, so D is correct.

Question 2

Question 2: 2. As shown in the figure, within the elastic limit, the spring with compression of $x$ is slowly st...

2. As shown in the figure, within the elastic limit, the spring with compression of $x$ is slowly stretched to an elongation of $x$, and the following statements about the change in elastic potential energy of the spring during this process are correct ( ![](/images/questions/phys-work-energy/image-002.jpg)

A. A. keep on decreasing
B. B. run
C. C. Decrease and then increase
D. D. Increase and then decrease

Answer

Answer: C

Solution: Because the spring is compressed at the beginning, before restoring the original length, the amount of deformation in the stretching process becomes smaller, according to the formula $E _ { \mathrm { p } } = \frac { 1 } { 2 } k x ^ { 2 }$ the elastic potential energy decreases, after restoring the amount of deformation, the amount of deformation in the stretching process becomes larger, according to the formula $E _ { \mathrm { p } } = \frac { 1 } { 2 } k x ^ { 2 }$ the elastic potential energy increases, so it is firstly decreasing and then increasing.

Question 3

Question 3: 4. The diagram shows the effect of a stroboscopic photo of an apple falling freely. The apple is loc...

4. The diagram shows the effect of a stroboscopic photo of an apple falling freely. The apple is located at the point $O$ when the first flash occurs, and the apple is located at the points $A , B , C$ when the second, third, and fourth flashes occur, respectively. It is known that $O A = x _ { 1 } , A B = x _ { 2 } , B C = x _ { 3 }$ and the flash period of the strobe is $T$. From the figure $x _ { 1 } < x _ { 2 } < x _ { 3 }$ , then the kinetic energy of the apple falling freely is ![](/images/questions/phys-work-energy/image-003.jpg)

A. A. larger and larger
B. B. remain unchanged
C. C. smaller and smaller
D. D. Getting smaller before getting bigger

Answer

Answer: A

Solution: As can be seen from the figure, the distance the apple falls in equal time becomes larger and larger, so the speed becomes larger and larger, and the kinetic energy of the apple becomes larger and larger. Therefore, choose A.

Question 4

Question 4: 5. As shown in the figure, a student with a mass of 50 kg is doing sit-ups. If the mass of the stude...

5. As shown in the figure, a student with a mass of 50 kg is doing sit-ups. If the mass of the student's upper body is about $\frac { 3 } { 5 }$ of the mass of the whole body, she did 50 sit-ups in 1 min, each time the center of gravity of the upper body rises to a distance of 0.3 m, then each time she sits up to overcome the force of gravity to do the work of $W$ and the power of the sit-up $P$ is about (). FORMULA_2]] is about ( ) ![](/images/questions/phys-work-energy/image-004.jpg)

A. A. $W = 150 \mathrm {~J} , P = 125 \mathrm {~W}$
B. B. $W = 150 \mathrm {~J} , P = 250 \mathrm {~W}$
C. C. $W = 90 \mathrm {~J} , P = 75 \mathrm {~W}$
D. D. $W = 90 \mathrm {~J} , P = 150 \mathrm {~W}$

Answer

Answer: C

Solution: Each time the center of gravity of her upper body rises to a height of 0.3 m, then the work done by her to overcome gravity is $$ W = \frac { 3 } { 5 } m g h = 90 \mathrm {~J} $$ The time for each sit-up is $t = 1.2 \mathrm {~s}$ , and the corresponding average power is approximately $$ P = \frac { W } { t } = 75 \mathrm {~W} $$ Therefore, ABD is wrong; C is correct.

Question 5

Question 5: 6. The following physical quantities must change during the process of curved motion of the object i...

6. The following physical quantities must change during the process of curved motion of the object is

A. A. accelerations
B. B. kinetic
C. C. momentum
D. D. mechanical energy

Answer

Answer: C

Solution: A. An object can be subjected to a constant force in curved motion, such as parabolic motion, and the acceleration can be constant, so A is wrong. B. In the process of curved motion, the direction of speed must change, but the size does not necessarily change, and the kinetic energy does not necessarily change, such as uniform circular motion, so B is wrong. C. The direction of speed must change in curved motion, and the direction of momentum is the direction of speed, so C is correct. D. There is no direct relationship between the curved motion of an object and its mechanical energy, such as uniform circular motion in a horizontal plane, so D is wrong.

Question 6

Question 6: 7. As shown in the figure, a mass of $m$ of the slider with an initial speed of ${ } ^ { V _ { 0 } }...

7. As shown in the figure, a mass of $m$ of the slider with an initial speed of ${ } ^ { V _ { 0 } }$ from the bottom of the incline along the incline upward movement, known as the kinetic friction factor between the slider and the incline of $\mu$, the slider rises the maximum height of $h$, then the slider from the bottom to the highest point in the process of sliding to the highest point. FORMULA_3]], then during the slide of the slider from the bottom of the incline to the highest point ![](/images/questions/phys-work-energy/image-005.jpg)

A. A. The work done on the slider by the incline support force is $\frac { m g h } { \tan \theta }$
B. B. The work done by friction on the slider is $\frac { \mu m g h } { \tan \theta }$
C. C. The work done by gravity on the slider is $m g h$
D. D. The work done on the slider by the combined external force is $- \frac { 1 } { 2 } m v _ { 0 } ^ { 2 }$

Answer

Answer: D

Solution: A. In the process of sliding up the small object, the support force of the inclined plane is always perpendicular to the speed, and does not do work on the slider, so A is wrong; B. According to the definition of work, the work done by the friction force on the slider is $$ W = - \mu m g \cos \theta \cdot \frac { h } { \sin \theta } = - \mu m g \frac { h } { \tan \theta } $$ Therefore, B is wrong; C. According to the characteristics of work done by gravity, it can be seen that the work done by gravity on the slider is $- m g h$, so C is wrong; D. rise to the highest point of zero speed, by the theorem of kinetic energy can be known that the work done by the combined external force on the slider is $$ W _ { \text {合 } } = \Delta E _ { \mathrm { k } } = 0 - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } = - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } $$ Therefore, D is correct.

Question 7

Question 7: 8. The small iron ball firm straight up and back to the throwing point in the process, then about th...

8. The small iron ball firm straight up and back to the throwing point in the process, then about the work done by gravity is correct ( )

A. A. the gravitational force does negative work
B. B. gravity does not do any work
C. C. the gravitational force is doing positive work
D. D. Gravity does negative work before positive work.

Answer

Answer: D

Solution: During the rise of the ball, gravity is downward and displacement is upward, then gravity does negative work; during the fall, gravity is downward and displacement is downward, then gravity does positive work, that is, the whole process of gravity first does negative work, and then does positive work.

Question 8

Question 8: 9. As shown in the figure, the ball is located on a smooth inclined plane, the inclined plane is loc...

9. As shown in the figure, the ball is located on a smooth inclined plane, the inclined plane is located on the smooth horizontal ground. From the ground, in the process of the ball sliding down the incline, the incline on the ball's force () ![](/images/questions/phys-work-energy/image-006.jpg)

A. A. Perpendicular to the contact surface, negative work
B. B. Perpendicular to the contact surface, doing zero work
C. C. Not perpendicular to the contact surface, does zero work
D. D. Not perpendicular to the contact surface, work not zero

Answer

Answer: A

Solution: The force of the inclined plane on the ball is elastic and perpendicular to the contact surface; because the ground is smooth, according to the conservation of momentum, it can be seen that the inclined plane will move horizontally to the left, the angle between the support force and the direction of the displacement of the ball is greater than $90 ^ { \circ }$, and the support force has done negative work on the ball, BCD is wrong A correct

Question 9

Question 9: 10. An object moves in a straight line with variable speed. At ${ } ^ { t _ { 1 } }$ the rate is $v$...

10. An object moves in a straight line with variable speed. At ${ } ^ { t _ { 1 } }$ the rate is $v$, at ${ } ^ { t _ { 2 } }$ the rate is $n v$ and at ${ } ^ { t _ { 2 } }$ the kinetic energy of the object is the same as that at ${ } ^ { t _ { 1 } }$. The kinetic energy of the object at the ${ } ^ { t _ { 1 } }$ moment is the ${ } ^ { t _ { 1 } }$ moment.

A. A. $n$ times
B. B. $\frac { n } { 2 }$ times
C. C. $n ^ { 2 }$ times
D. D. $\frac { n ^ { 2 } } { 4 }$ times

Answer

Answer: C

Solution: From $E _ { k } = \frac { 1 } { 2 } m v ^ { 2 }$, the velocity becomes $n$ times, then the kinetic energy becomes $n ^ { 2 }$ times, C is correct;

Question 10

Question 10: 11. As shown in the figure, a bullet with a horizontal speed into the original stationary block plac...

11. As shown in the figure, a bullet with a horizontal speed into the original stationary block placed on a smooth horizontal surface, and stay in the block, in the process of the bullet drilled into the depth of the block for $d$, the displacement of the block for $l$, the size of friction between the block and the bullet for $F$, then ( ), then $F$, then ( ), then ( ), then ( ), then ( ), then ( ), then ( ), then ( ), then ( ), then ( ) FORMULA_2]], then ( ) ![](/images/questions/phys-work-energy/image-007.jpg)

A. A. $F$ The work done on the block is $F l$
B. B. $F$ The work done on the block is $F ( l + d )$
C. C. $F$ does work on the bullet as $- F d$
D. D. $F$ does work on the bullet as $F ( l + d )$

Answer

Answer: A

Solution: AB. The displacement of the block is $l$, and from $W = F l \cos \alpha$ The work done on the block by $F$ is $W = F l$. A is correct; B is incorrect; CD. The displacement of the bullet is $l + d$, and the direction of the friction force of the block on the bullet is opposite to the direction of the displacement, so the friction force of the block on the bullet is $l + d$. The friction force on the bullet is negative, i.e. $W = - F ( l + d )$ CD Error.

Question 11

Question 11: 12. As shown in the figure, in the horizontal table on the $A$ point there is a mass of $m$ of the o...

12. As shown in the figure, in the horizontal table on the $A$ point there is a mass of $m$ of the object to the initial velocity $v _ { 0 }$ was thrown, regardless of air resistance, when it reaches the $B$ point, its kinetic energy is () ![](/images/questions/phys-work-energy/image-008.jpg)

A. A. $\frac { 1 } { 2 } m v _ { 0 } ^ { 2 } + m g H$
B. B. $\frac { 1 } { 2 } m v _ { 0 } ^ { 2 } + m g h$
C. C. $m g H - m g h$
D. D. $\frac { 1 } { 2 } m v _ { 0 } ^ { 2 } + m g ( H - h )$

Answer

Answer: B

Solution: Not counting air resistance, only gravity does the work, from $A$ to $B$. By the kinetic energy theorem $$ \begin{aligned} & E _ { K B } - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } = m g h \\ & E _ { K B } = \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } + m g h \end{aligned} $$

Question 12

Question 12: 13. As shown in the figure, a straight wooden board can be rotated in a straight plane around a hori...

13. As shown in the figure, a straight wooden board can be rotated in a straight plane around a horizontal axis passing through the point $O$, and there is a small wooden block on the board, the right end of which is subjected to a straight upward force $F$, and it is slowly rotated from position A to position B. The small block has not slipped relative to the board, so during this process ( ), the board has not slipped. The block does not slide relative to the board, then during this process ( ) [IMAGE_2]]

A. A. The friction force on the block does positive work first and then negative work.
B. B. The friction force on the small block always does positive work on the small block
C. C. The supporting force on the small block does positive work on the small block
D. D. The supporting force on the small wooden block always does no work on the small wooden block

Answer

Answer: C

Solution: AB . The friction force on the block is always perpendicular to the direction of motion, so no work is done, so AB is wrong; CD.The support force on the small wooden block is always perpendicular to the direction of the water plate and is always in the same direction as the motion of the small wooden block, so the support force on the small wooden block always does positive work on the small wooden block, so C is correct and D is wrong.

Question 13

Question 13: 14. Throw a ball with mass $m$ upward with a certain initial velocity, the maximum height of the bal...

14. Throw a ball with mass $m$ upward with a certain initial velocity, the maximum height of the ball is $h$, and the magnitude of air resistance is $F$, then from the process of throwing to the process of falling back to the point of throwing, the work of the ball and the total work of all the external forces on the ball are ( ) respectively. The work done by the air resistance on the ball and the total work done by all the external forces on the ball are ( ) $- 2 F h$

A. A. $0 \quad 0$
B. B. -Fh mgh
C. C. Fh $m g h$
D. D. $- 2 F h$

Answer

Answer: D

Solution: The work done by the resistance during the whole process is $$ W _ { \mathrm { f } } = - 2 h F $$ Throughout the process, the ball is only subjected to gravity and resistance, and back to the initial position, then gravity does not do work, so the total work done by the combined external force is the work done by the resistance, that is - $2 h F$. Therefore, ABC is wrong and D is correct.

Question 14

Question 14: 15. As shown in the figure, the mass of the same two balls A, B, respectively, with a long $l$ and $...

15. As shown in the figure, the mass of the same two balls A, B, respectively, with a long $l$ and $2 l$ of light rope hanging at different heights on the ceiling, now pull up the two balls so that the rope straightened to the horizontal state, and then released by the static, the two small balls during the movement of the location of the lowest point in the same horizontal plane, to the small balls A began to release the position of the horizontal plane of gravity for the zero potential energy reference surface, when the two small balls to reach the lowest position ( ) ![](/images/questions/phys-work-energy/image-010.jpg)

A. A. The two balls are moving with equal linear velocities
B. B. The two balls are moving with equal angular velocities
C. C. Taking the same horizontal plane as the point of zero potential energy, the mechanical energies of the two balls are equal
D. D. The pull of the rope on the two balls is equal

Answer

Answer: D

Solution: A. The velocity when reaching the lowest point is $v$ : with $m g l = \frac { 1 } { 2 } m v ^ { 2 }$, then $$ v = \sqrt { 2 g l } $$ Because the rope lengths are different, the linear velocities are not equal, so A is wrong; B. When the two balls reach the lowest position, the angular velocity of the two ball movement $$ \omega = \frac { v } { r } = \frac { \sqrt { 2 g l } } { l } = \sqrt { \frac { 2 g } { l } } $$ Because the length of the rope is different, the angular velocity is different, so B is wrong; C. The initial positions of the two balls are different, the gravitational potential energy is different, the kinetic energy is the same, then the mechanical energy is different, so C is wrong; D. The tension of the rope is $F$, and at the lowest point: $F - m g = m \frac { v ^ { 2 } } { l }$, solve $$ F = 3 m g $$ is independent of the rope length, so D is correct.

Question 15

Question 15: 16. In the derivation of "uniformly variable linear motion displacement formula", the entire process...

16. In the derivation of "uniformly variable linear motion displacement formula", the entire process of motion is divided into a number of small segments, each small segment of the approximation of uniform linear motion, and then the displacement of the small segments to calculate the displacement of the entire process of summing up the displacement of the entire process, physics in this method is called the "micro element method This method is called the "micro-element method" in physics. The following examples apply to this method of thinking ( )

A. A. When calculating the gravitational force between objects, if the size of the object is relatively small, the object can be regarded as a prime point
B. B. In the process of investigating the expression for elastic potential energy, the process of stretching a spring is divided into many small segments, and within each segment the elastic force of the spring is considered to be a constant force, and then the algebraic sum of the work done by the elastic force in each segment is added to find the value of the work done by the elastic force for the entire process
C. C. Investigating Newton's second law by controlling the mass of an object to be constant and investigating the relationship between the acceleration of an object and the force on it
D. D. When finding the combined force of two forces, if the effect of one force is the same as the effect of two forces, this force is the combined force of those two forces

Answer

Answer: B

Solution: A. In the calculation of the gravitational force between objects, if the size of the object is relatively small, the object can be regarded as a mass, the use of idealized models, not the method of microelements, so A does not meet the meaning of the question; B. In the process of investigating the expression of elastic potential energy, the process of stretching the spring is divided into a number of small segments, in each small segment of the spring is considered a constant force, and then each small segment of the elasticity of the work done by the algebraic sum of the entire process of elasticity of the work done by the numerical value, the use of microelementary, so B meet the meaning of the question; C. Newton's second law of the process, control the mass of the object is unchanged, the study of the relationship between the acceleration of the object and the force of the use of the controlled variable method, so C does not meet the meaning of the question; D. When looking for the combined force of two forces, if the effect of a force is the same as the effect of two forces, this force is the combined force of the two forces, using the method of equivalence, so D does not meet the meaning of the question.

Question 16

Question 16: 17. As shown in the figure, an object in 10 s along the horizontal road to the right of the movement...

17. As shown in the figure, an object in 10 s along the horizontal road to the right of the movement of 1 m, in the process the object has been subjected to a size of 10 N, the direction of the direction diagonally to the upper left and with the horizontal direction of the $60 ^ { \circ }$ of the role of the tension $F$. Then the work done by the tension $F$ on the object during this time is () ![](/images/questions/phys-work-energy/image-011.jpg)

A. A. 10 J
B. B. - 10 J
C. C. 5 J
D. D. - 5 J

Answer

Answer: D

Solution: Calculate the expression for work done according to the constant force $$ W = F L \cos \alpha $$ it is possible to obtain $$ W = 10 \times 1 \times \cos 120 ^ { \circ } \mathrm { J } = - 5 \mathrm {~J} $$

Question 17

Question 17: 18. As shown in the figure, with a line tethered ball in the smooth horizontal surface to do uniform...

18. As shown in the figure, with a line tethered ball in the smooth horizontal surface to do uniform circular motion, the radius of the circle is 1 m, the ball's mass is 0.1 kg, the line speed ${ } ^ { V = 1 \mathrm {~m} / \mathrm { s } }$, the ball from the $A$ point of the movement to the $B$ point exactly half of the circumference, then in the this motion the work done by the tension of the line is ( ) [IMAGE_3]]

A. A. 0
B. B. 0.1 J
C. C. 0.314 J
D. D. indeterminate

Answer

Answer: A

Solution: During the movement of the ball, the direction of the tension of the rope is always perpendicular to the direction of the speed of the ball, so the work done by the tension is 0, so A is correct;

Question 18

Question 18: 19. As shown in the figure, the elevator mass is $M$, it is placed on the horizontal floor of a mass...

19. As shown in the figure, the elevator mass is $M$, it is placed on the horizontal floor of a mass of $m$ of the object, the elevator in the steel cable under the action of the tension by the static start of the straight upward accelerated motion. When the rise height is $H$, the elevator's speed reaches $v$, then in this process, the following statements are correct ( ) ![](/images/questions/phys-work-energy/image-013.jpg)

A. A. The work done by the elevator to support the object is equal to $\frac { 1 } { 2 } m v ^ { 2 }$
B. B. The work done by the elevator to support the object is greater than $\frac { 1 } { 2 } m v ^ { 2 }$
C. C. The work done by the tension in the cable is equal to $\frac { 1 } { 2 } m v ^ { 2 } + M g H$
D. D. The work done by the tension in the cable is less than $\frac { 1 } { 2 } m v ^ { 2 } + M g H$

Answer

Answer: B

Solution: A. The sum of the positive work done by the support of the ladder on the object and the negative work done by the object's gravity is equal to $\frac { 1 } { 2 } m v ^ { 2 }$, so A is wrong; B. From the analysis of item A, the work done by the elevator's supporting force on the object is equal to the work done by the object overcoming the force of gravity plus $\frac { 1 } { 2 } m v ^ { 2 }$, so B is correct; CD . By the kinetic energy theorem [BLOCK_FORMULA_0]] The solution is $$ W _ { \text {钢索 } } = ( m + M ) g H + \frac { 1 } { 2 } ( m + M ) v ^ { 2 } $$ Therefore, CD is wrong.

Question 19

Question 19: 20. with a tension force F will be a weight of 10 N of the object raised at uniform speed 1 m, as sh...

20. with a tension force F will be a weight of 10 N of the object raised at uniform speed 1 m, as shown in the figure, in the process, the following statements are wrong ( ) ![](/images/questions/phys-work-energy/image-014.jpg)

A. A. The work done on the object by the pulling force F is 10 J.
B. B. The work done on the object by the combined force is 10 J
C. C. The work done by the object to overcome gravity is 10 J
D. D. The gravitational potential energy of the object increases by 10 J

Answer

Answer: B

Solution: A. Because the object rises at a constant speed, then $F = G = 10 \mathrm {~N}$, then the tension force does work $W _ { F } = F h = 10 \times 1 \mathrm {~J} = 10 \mathrm {~J}$, so $A$ is correct; B. The object rises at a constant speed and the magnitude of the combined force is zero, so the work done by the combined force is zero, so B is wrong; CD. gravity does work $W _ { G } = - m g h = - 10 \mathrm {~J}$, know that the object overcomes gravity to do work 10 J, the gravitational potential energy increased by 10 J, so $\mathrm { C } , D$ is correct; The statement is wrong, so choose B.

Question 20

Question 20: 22. As shown in the figure, Xiaoming pulls a wooden box with a light rope that makes an angle of $\t...

22. As shown in the figure, Xiaoming pulls a wooden box with a light rope that makes an angle of $\theta$ with the horizontal, the tension in the rope is $F$, and the wooden box moves along the horizontal ground to the right for a distance of $l$. The kinetic friction factor between the crate and the ground is $\mu$, the mass of the crate is $m$, and the crate is subjected to a friction factor of ![](/images/questions/phys-work-energy/image-015.jpg). ![](/images/questions/phys-work-energy/image-015.jpg)

A. A. The work done by the support force is $( m g - F \sin \theta )$
B. B. The work done by gravity is $m g l$
C. C. The work done by the pulling force is $F l \cos \theta$
D. D. The work done by sliding friction is $- \mu \mathrm { mgl }$

Answer

Answer: C

Solution: A. The direction of the supporting force is straight upward, always perpendicular to the direction of the movement of the wooden box, so the supporting force does not do work, so A is wrong; B. The direction of gravity is straight down, always perpendicular to the direction of the movement of the wooden box, so gravity does not do work, so B error; C. The angle between the direction of the pulling force and the direction of the movement of the wooden box is $\theta$, so the pulling force does work. $$ W _ { \mathrm { F } } = F l \cos \theta $$ ## Therefore, C is correct; D. The force on the wooden box can be analyzed to obtain the sliding friction force of $$ F _ { \mathrm { f } } = \mu F _ { \mathrm { N } } = \mu ( m g - F \sin \theta ) $$ Then the work done by friction $$ W _ { F _ { \mathrm { f } } } = - F _ { \mathrm { f } } l = - \mu ( m g - F \sin \theta ) l $$ Therefore, D is wrong.

Question 21

Question 21: A body of mass 2 kg is in free fall and lands in 2 s. The acceleration due to gravity is $\mathrm { ...

A body of mass 2 kg is in free fall and lands in 2 s. The acceleration due to gravity is $\mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$. The acceleration due to gravity is $\mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$. The following statements about the work done by gravity and the power are correct.

A. A. The average power of gravity during the fall is 200 W
B. B. The work done by gravity during the fall is 200 J
C. C. The kinetic energy of the object at the instant before it hits the ground is 200 J.
D. D. The instantaneous power of gravity at the moment before hitting the ground is 200 W

Answer

Answer: A

Solution: AB. Work done by gravity $$ P = m g h = 2 \times 10 \times 20 \mathrm {~J} = 400 \mathrm {~J} $$ Average power $$ \bar { P } = \frac { m g h } { t } = \frac { 2 \times 10 \times 20 } { 2 } \mathrm {~W} = 200 \mathrm {~W} $$ A is correct, B is incorrect; CD. ground rate at landing $$ v = g t = 10 \times 2 \mathrm {~m} / \mathrm { s } = 20 \mathrm {~m} / \mathrm { s } $$ The kinetic energy of an object at the instant before landing is $$ P = \frac { 1 } { 2 } m v ^ { 2 } = 400 \mathrm {~J} $$ The instantaneous power of gravity $$ P = m g v = 20 \times 20 \mathrm {~W} = 400 \mathrm {~W} $$ Therefore, C is wrong and D is wrong.

Question 22

Question 22: 24. The following is a correct description of the understanding of work.

24. The following is a correct description of the understanding of work.

A. A. Because work has a positive and negative value, work is a vector quantity
B. B. When the action force does positive work, the reaction force must do negative work
C. C. The mechanical energy of an object must be conserved when the work done by the combined external force on the object is zero
D. D. Static friction may do positive work, negative work, or no work at all

Answer

Answer: D

Solution: A. Work is positive and negative, but work is a scalar quantity, so A error; B. A pair of interacting forces to do work, can appear to do positive work, do negative work, a do positive work, another do negative work or a do work, a do not do work and so on various cases, so B error; C. In the role of the external force upward uniform velocity, the work of the combined force is zero, the kinetic energy is unchanged, but the gravitational potential energy increases, the mechanical energy increases, so C error; D. The direction of static friction and the direction of the relative motion of the object trend in the opposite direction, but with the direction of motion can be the same, can also be the opposite, but also perpendicular to the direction of motion, so static friction can do positive work, can also do negative work, but also can not do the work, so D is correct.

Question 23

Question 23: When an object is thrown straight upwards, the object rises to a maximum height of $h$, and if the m...

When an object is thrown straight upwards, the object rises to a maximum height of $h$, and if the mass of the object is $m$ and the air resistance is constant at $f$, then during the entire process from the moment it is thrown to the moment it falls back to the ground (the object is thrown upwards to the moment it falls back to the ground).

A. A. The work done by air resistance is zero
B. B. The work done by air resistance is $2 f h$
C. C. The work done by gravity is zero
D. D. The work done by gravity is $2 m g h$

Answer

Answer: C

Solution: AB . During the ascent, the work done by air resistance is $- f h$ and during the descent, the work done by air resistance is - $f h$, then the work done by air resistance during the whole process is $- 2 f h , \mathrm { AB }$ Error; CD. the object was thrown to fall back to the throw point in the whole process, the beginning and end of the same position, the height difference is zero, so the work done by gravity is zero, C is correct, D is wrong.

Question 24

Question 24: 26. A wooden block has a pulley at the front end, one end of the rope is tied to a fixed place on th...

26. A wooden block has a pulley at the front end, one end of the rope is tied to a fixed place on the right, and the other end is pulled through the pulley with a constant force F to keep the angle between the two strands of the rope $\theta$ unchanged, as shown in the figure, when pulling the rope to make the wooden block move forward for s, the force F on the wooden block to do the work (not counting the weight of the rope and friction) is ( ) ![](/images/questions/phys-work-energy/image-016.jpg)

A. A. Fscos $\theta$
B. B. Fs $( 1 + \cos \theta )$
C. C. 2 Fscos $\theta$
D. D. 2 Fs

Answer

Answer: B

Solution: When pulling the rope, the tension in both strands is $F$ and they both do work on the object, according to the expression for work done by a constant force: $\mathrm { W } = \mathrm { FS } \cos \theta$, we get: the tension force $F$ that is directed diagonally upward to the right does work as: $W _ { 1 } = \mathrm { FS } \cos \theta$ The work done by the horizontal rightward pulling force is: $W _ { 2 } = \mathrm { FS }$ Therefore the work done on the object is $W = W _ { 1 } + W _ { 2 } = F S \cos \theta + F S = F S ( 1 + \cos \theta ) \equiv$

Question 25

Question 25: 27. A high-speed car due to brake failure, the driver will be the car into the outside of the ramp-t...

27. A high-speed car due to brake failure, the driver will be the car into the outside of the ramp-type refuge lane, as shown in the figure. In the process of the car rushing up the ramp to do deceleration, the following statements are correct ( ) ![](/images/questions/phys-work-energy/image-017.jpg)

A. A. The gravity of the car does positive work
B. B. The gravity of the car does negative work
C. C. The gravitational potential energy of the car decreases
D. D. The gravitational potential energy of the car remains constant

Answer

Answer: B

Solution: AB . The car's gravity does negative work, so A is wrong; B is correct; CD.The car's gravitational potential energy increases, so CD is wrong;

Question 26

Question 26: 28. With regard to geostationary satellites, the following statements are incorrect ( ).

28. With regard to geostationary satellites, the following statements are incorrect ( ).

A. A. Launch velocity greater than the first cosmic velocity
B. B. Orbiting at less than the first cosmic velocity
C. C. The period of operation is certain, but the height above the ground may be different.
D. D. Different geostationary satellites may have different kinetic energies.

Answer

Answer: C

Solution: A. The first cosmic velocity is the smallest launch speed, then the speed when launching a geostationary satellite is greater than the first cosmic velocity, option A is correct and does not meet the meaning of the question; B.The first cosmic velocity is the maximum encircling velocity, then the operating speed at the launch of the geostationary satellite is less than the first cosmic velocity, option B is correct and does not meet the meaning of the question; C. All geostationary satellites have a certain period of operation, are 24 h, according to the $$ G \frac { M m } { r ^ { 2 } } = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } r $$ According to $$ G \frac { M m } { r ^ { 2 } } = m \frac { 4 \pi ^ { 2 } } { T ^ { 2 } } r $$, the altitude from the ground is the same, option C is wrong, consistent with the meaning of the question; D. The mass of different geostationary satellites may be different, the kinetic energy may be different, option D is correct, does not meet the meaning of the question.

Question 27

Question 27: 29. mass of $m$ objects along the bottom of the length are $L$, the angle of inclination is differen...

29. mass of $m$ objects along the bottom of the length are $L$, the angle of inclination is different $a , b , c$ three slopes from the top slide down, as shown in the figure, the object and the slopes of the sliding friction factor is the same, $a , b , c$ three slopes and the horizontal plane angle is the relationship between the $a , b , c$ three slopes and the horizontal plane is the same, the $a , b , c$ three slopes and the horizontal plane is the relationship between the slopes and the slopes of the slopes of the slopes. 3]] The relationship between the angles of the three inclined surfaces and the horizontal plane is $\theta _ { 1 } > \theta _ { 2 } > \theta _ { 3 }$, the object from $a , b , c$ the top of the three slopes to the bottom of the process, the friction force to do work are $W _ { 1 } , W _ { 2 }$ and $W _ { 3 }$, then they are related to the relationship between the ( ) ![](/images/questions/phys-work-energy/image-018.jpg)

A. A. $W _ { 1 } > W _ { 2 } > W _ { 3 }$
B. B. $W _ { 1 } = W _ { 3 } > W _ { 2 }$
C. C. $W _ { 1 } > W _ { 2 } = W _ { 3 }$
D. D. $W _ { 1 } = W _ { 2 } = W _ { 3 }$

Answer

Answer: D

Solution: As seen in the picture ![](/images/questions/phys-work-energy/image-019.jpg) The friction force is equal to $$ f = \mu F _ { \mathrm { N } } = \mu m g \cos \theta $$ The length of the bottom edge is $L$ and the displacement traveled by the object is $x$, the $$ x = \frac { L } { \cos \theta } $$ From the formula for work done $$ W _ { f } = - f x = - \mu m g L $$ Therefore, the work done by friction is related to $\mu , m , L$, not to the angle.

Question 28

Question 28: 30. Manned flight bag is a single person low altitude flight device, as shown in the figure, its eng...

30. Manned flight bag is a single person low altitude flight device, as shown in the figure, its engine uses gasoline as fuel to provide power, can be vertical takeoff and landing can also be fast forward, if the flight bag (including the person) in the firm straight direction in the process of rising at a constant speed (air resistance can not be ignored), the following statements are correct () ![](/images/questions/phys-work-energy/image-020.jpg)

A. A. The engine does no work on the flight bag
B. B. Positive work done by the gravity of the flight bag
C. C. The kinetic energy of the flight pack is constant
D. D. The mechanical energy of the flight bag remains constant

Answer

Answer: C

Solution: A. In the process of the flight bag (including people) rising vertically at a constant speed, the power of the engine is upward, then the engine does positive work on the flight bag, so A is wrong; B. The height rises, the gravity of the flight bag does negative work, so B is wrong; C. Flight packages (including people) in the direction of the firm straight up at uniform speed, so the kinetic energy of the flight packages remain unchanged, so C is correct; D. The kinetic energy of the flight bag is unchanged in the process of rising, and the gravitational potential energy becomes larger, so the mechanical energy becomes larger, so D is wrong.

Question 29

Question 29: 31. The following statements are correct

31. The following statements are correct

A. A. The acceleration of an object in circular motion must be directed toward the center of the circle
B. B. As long as the sum of the external forces on the system is zero, the momentum of the system is conserved and the mechanical energy of the system is also conserved
C. C. Throwing an object in different directions at the same location on the earth, the object has the same rate of change of velocity, regardless of air resistance
D. D. A point in simple harmonic motion passes through the same point successively, and the restoring force, velocity, acceleration, and displacement must be the same.

Answer

Answer: C

Solution: A. Uniform circular motion of the object, the acceleration points to the center of the circle, variable-speed circular motion has tangential acceleration to change the size of the speed, so the A option is wrong; B. The combined external force is zero, momentum conservation, but if there is interaction within the system of sliding friction, will be accompanied by friction heat, mechanical energy is not conserved, the B option is wrong; C. rate of change of speed is equal to acceleration, regardless of air resistance, throwing motion only by gravity, acceleration is the acceleration of gravity, the same place, the acceleration of gravity is the same, so the C option is correct; D. Simple harmonic motion of the point successively through the same point, the direction of the speed is different, so the D option is wrong.

Question 30

Question 30: 32. As shown in the figure for the investigation of the elastic potential energy of the spring with ...

32. As shown in the figure for the investigation of the elastic potential energy of the spring with which factors related to the experimental setup, with a downward force slowly pushed the ball to compress the spring, after the withdrawal of the force, the ball from the firm straight round tube upward ejection. Not counting friction and air resistance, the spring deformation is always within the elastic limit, the ## ||

A. A. By increasing the compression of the spring only, the height of the ball rises unchanged
B. B. By increasing the compression of the spring only, the height of the ball rises increases
C. C. By increasing only the coefficient of strength of the spring, the height at which the ball rises remains unchanged
D. D. By increasing only the coefficient of strength of the spring, the height of the ball rises decreases

Answer

Answer: B

Solution: Let the initial compression of the spring be $x$, the initial spring coefficient of strength be $k$, the height of the ball in the initial state is $h$, and the mass of the ball is $m$, then the spring will be released by an external force. During the process from the release of the spring to the rise of the ball to its maximum height, the ball is subjected to a force of Gravity and elastic force together, its mechanical energy increment is equal to the work done by the elastic force, the size and the initial elastic potential energy $E _ { \mathrm { p } } = \frac { 1 } { 2 } k x ^ { 2 }$ is equal, that is $$ m g h = \frac { 1 } { 2 } k x ^ { 2 } $$ The ball rises to a height of $$ h = \frac { k x ^ { 2 } } { 2 m g } $$ From the above equation, if only the compression of the spring is increased the height of the rise of the ball increases, and only the coefficient of strength of the spring is increased, the height of the rise of the ball increases.

Question 31

Question 31: 33. An object of mass 1 kg slides on a horizontal surface, its kinetic energy varies with displaceme...

33. An object of mass 1 kg slides on a horizontal surface, its kinetic energy varies with displacement as shown in the figure, take $\mathrm { g } = 10 \mathrm { m } / \mathrm { s } ^ { 2 }$, then the object slides for the duration of () ![](/images/questions/phys-work-energy/image-021.jpg)

A. A. 2 s
B. B. 3 s
C. C. 4 s
D. D. 5 s

Answer

Answer: D

Solution: According to the kinetic energy theorem $\mathrm { W } = \mathrm { E } _ { \mathrm { k } 2 } - \mathrm { E } _ { \mathrm { k } 1 }$, from the diagram, the kinetic energy is decreasing, which means that the resistance is doing negative work on the object, so there is $- \mathrm { Fs } = 0 - \mathrm { E } _ { \mathrm { k } }$ $\mathrm { F } = \frac { E _ { K } } { s } = \frac { 50 } { 25 } \mathrm {~N} =$"2 "N Acceleration $\mathrm { a } = \frac { F } { m } = \frac { 2 } { 1 } \mathrm {~m} / \mathrm { s } ^ { 2 } = " 2 " \mathrm {~m} / \mathrm { s } ^ { 2 }$ From the kinematic equation $\mathrm { s } = \frac { 1 } { 2 } \mathrm { at } ^ { 2 }$ we get $: \mathrm { t } = \sqrt { \frac { 2 s } { a } } = \sqrt { \frac { 2 \times 25 } { 2 } } \mathrm {~s} =$" 5 " s. Idea Analysis: The kinetic energy of the object is decreasing with time, so the resistance is doing negative work, according to the kinetic energy theorem $\mathrm { W } = \mathrm { E } _ { \mathrm { k } 2 } - \mathrm { E } _ { \mathrm { k } 1 } , ~ \mathrm { a } = \frac { F } { m }$ , $\frac { 1 } { 2 } \mathrm { at } ^ { 2 }$ analyze that Test comment: This question examines the use of the theorem of kinetic energy, combined with the given image analysis to solve the problem

Question 32

Question 32: 34. new coronavirus caused by the pneumonia epidemic touched the hearts of hundreds of millions of p...

34. new coronavirus caused by the pneumonia epidemic touched the hearts of hundreds of millions of people in the country, a party has a problem, eight support, when Wuhan, a variety of items in emergency, the people of the country have been generous, donations, donations to support Wuhan. 28 January afternoon, the first batch of Shandong Shouguang Wuhan aid more than 350 tons of more than 20 kinds of fresh high-quality vegetables loaded, as shown in Figure A, loaded with vegetables along a section of a straight highway trucks slowly Driving out, its speed $v$ with the displacement $x$ of the relationship between the changes as shown in Figure B. The truck will be able to drive along a straight highway. The truck ![](/images/questions/phys-work-energy/image-022.jpg) Figure A ![](/images/questions/phys-work-energy/image-023.jpg) Figure B

A. A. travel in a straight line with uniform acceleration
B. B. Same amount of change in velocity in the same amount of time
C. C. Increasing change in velocity in the same amount of time
D. D. The work done by the combined external force is the same for the same displacement.

Answer

Answer: C

Solution: ABC. Because $$ \frac { \Delta v } { \Delta x } = \frac { \Delta v } { \Delta t } \cdot \frac { \Delta t } { \Delta x } = \frac { a } { v } $$ The slope of the graph of $v - x$ indicates $k = \frac { a } { v }$, the speed of the car increases, then the acceleration $a$ increases, and it is in accelerated motion with an increasing acceleration, according to $$ \Delta v = a \Delta t $$ According to $$ \Delta v = a \Delta t $$, the change of speed in the same time is getting bigger and bigger, option AB is wrong, C is correct; D. From the figure, we can see that the increase of velocity in the same displacement is the same, according to $$ \Delta v = a \Delta t $$. $$ W _ { \text {合 } } = \frac { 1 } { 2 } m v ^ { 2 } - \frac { 1 } { 2 } m v ^ { \prime 2 } = \frac { 1 } { 2 } m \left( v + v ^ { \prime } \right) \Delta v $$ D. The amount of work done by the combined external force in the same displacement is greater and greater, option D is wrong.

Question 33

Question 33: 35. December 11, 2020 Lianhuaiyang Town Railway line opened to traffic, realizing the dream of Yangz...

35. December 11, 2020 Lianhuaiyang Town Railway line opened to traffic, realizing the dream of Yangzhou all counties and districts by train. As shown in the figure, a train mass of $m$ of the moving car, initial speed of $v _ { 0 }$, with constant power $P$ in the flat track movement, by the time $t$ to reach the maximum speed of this power [[]], let the moving car traveling process is subjected to $\mathrm { v } _ { \mathrm { m } }$. INLINE_FORMULA_4]], and the resistance $F$ remains constant. In the time $t$, the moving car is traveling with ![](/images/questions/phys-work-energy/image-024.jpg) ![](/images/questions/phys-work-energy/image-024.jpg)

A. A. travel in a straight line with uniform acceleration
B. B. Gradual increase in acceleration
C. C. Power of traction $P = \mathrm { Fv } _ { \mathrm { m } }$
D. D. Work done by traction $W = \frac { m v _ { \mathrm { m } } ^ { 2 } - m v _ { 0 } ^ { 2 } } { 2 }$

Answer

Answer: C

Solution: AB. The moving car is moving with constant power according to $$ P _ { \text {额 } } = F _ { \text {奔 } } V $$ According to $$ P _ { \text {额 } } = F _ { \text {奔 } } V $$, as the speed increases, the traction force decreases, the combined force decreases, and the acceleration decreases, so the train moves in a straight line with decreasing acceleration, and AB is wrong; C. When the size of the traction force and resistance is equal to the acceleration is zero, the speed to reach the maximum, that is, there is a $$ F _ { \text {军 } } = F $$ Then the power of the moving car $$ P = F _ { \text {奔 } } v _ { m } = F v _ { m } $$ C Correct; D. According to the theorem of kinetic energy the work done by the combined external force is $$ W + W _ { \text {阻力 } } = \frac { 1 } { 2 } m v _ { m } ^ { 2 } - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } $$ Then the work done by the traction force is $$ W = \frac { 1 } { 2 } m v _ { m } ^ { 2 } - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } - W _ { \text {阻力 } } $$ D Error.

Question 34

Question 34: 36. As shown in the figure, the length of $L$ light rod is fixed at one end of the mass of $m$ of th...

36. As shown in the figure, the length of $L$ light rod is fixed at one end of the mass of $m$ of the ball, the other end is fixed to the rotating axis $O$, now the ball in the firm plane to do the circular motion, $P$ for the highest point of the circle, if the ball passes through the lowest point of the circle, the size of the velocity is $\sqrt { \frac { 9 } { 2 } g L }$, ignoring frictional resistance and air resistance, then the following judgment is correct ([INLINE_FORMULA_4]]. The highest point of the circle, if the ball through the lowest point of the circle when the speed size is $\sqrt { \frac { 9 } { 2 } g L }$, ignore frictional resistance and air resistance, then the following judgment is correct (). ![](/images/questions/phys-work-energy/image-025.jpg)

A. A. The ball cannot reach the $P$ point.
B. B. The speed of the ball when it reaches the point $P$ is greater than $\sqrt { g L }$.
C. C. The ball can reach the point $P$ and is bounced upward by the light rod at the point $P$.
D. D. The ball can reach the point $P$ and is bounced downward by the light rod at the point $P$.

Answer

Answer: C

Solution: AB. From the lowest point to the highest point, according to the kinetic energy theorem we have $$ - m g 2 L = \frac { 1 } { 2 } m v _ { P } ^ { 2 } - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } $$ Bringing in the data gives the velocity at the highest point $$ v _ { P } = \frac { \sqrt { 2 g L } } { 2 } $$ For the light rod, it can provide both upward support and downward tension, so no matter how much speed, as long as it can reach the highest point can pass the highest point, so the ball $P$ can reach the highest point, but the speed is less than $\sqrt { g L }$, the option A, B is wrong. CD . The force at the $P$ point is $P$. $$ F + m g = m \frac { v _ { p } ^ { 2 } } { L } $$ can be obtained $$ F = - \frac { m g } { 2 } $$ It means that the support force of the rod is firmly upward, option C is correct, D is wrong . [EYE POINT] circular motion combined force to provide centripetal force, light rod model and thin line light rope is different, the light rod of the elastic force is not necessarily along the direction of the rod, can be along the rod to the outside, can also be along the rod to the inside, so circular motion as long as you can get to the highest point, you can pass. The combined force provides centripetal force, you can assume that the direction of the elastic force is positive in the same direction as the direction of gravity, and the calculation result, if negative, indicates that the direction is opposite to the direction of gravity.

Question 35

Question 35: 37. In the smooth horizontal plane, a slider mass $m = 2 \mathrm {~kg}$, in the horizontal plane by ...

37. In the smooth horizontal plane, a slider mass $m = 2 \mathrm {~kg}$, in the horizontal plane by the horizontal direction of the constant external force $F =$ 4 N (direction unknown) under the action of the action of the slider, as shown in the figure shows a section of the slider in the horizontal plane of the movement of the trajectory of the slider, slider over the $P , Q$ two points when the speed size is $v = 5 \mathrm {~m} / \mathrm { s }$. _2]] at two points, the velocity of the slider is $v = 5 \mathrm {~m} / \mathrm { s }$. The slider's velocity at $P$ is at the angle $P Q$ with the line $\alpha = 37 ^ { \circ }$ and $\sin 37 ^ { \circ } = 0.6$, then the following statement is correct ![](/images/questions/phys-work-energy/image-026.jpg)

A. A. The direction of the horizontal constant force $F$ makes an angle of $53 ^ { \circ }$ with the line of $P Q$.
B. B. Slider from $P$ to $Q$ in 6 s
C. C. The minimum velocity of the slider from $P$ to $Q$ is $5 \mathrm {~m} / \mathrm { s }$.
D. D. $P , Q$ The distance between the two points is 12 m

Answer

Answer: D

Solution: A. Let the direction of the horizontal force $F$ and $P Q$ line angle is $\beta$, the slider over the $P , Q$ two points of the speed of the size of the same, according to the theorem of kinetic energy to get $$ F x _ { P Q } \cos \beta = \Delta E _ { k } = 0 $$ Gain $$ \beta = 90 ^ { \circ } $$ That is, the horizontal constant force $F$ is perpendicular to the line of $P Q$ and points to the concave side of the trajectory, A is wrong; B. The velocity at $P$ is decomposed into two directions along the horizontal force $F$ and the vertical force $F$, and the slider is first uniformly decelerated and then uniformly accelerated along the horizontal force $F$. linear motion, there are $$ a = \frac { F } { m } = 2 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ When the velocity in $F$ direction is zero, the time is $$ v \sin 37 ^ { \circ } = a t $$ Solve for $$ t = 1.5 \mathrm {~s} $$ According to the symmetry, the time of the slider from $P$ to $Q$ is $$ t ^ { \prime } = 2 t = 3 \mathrm {~s} $$ B Error ; C. When the velocity in the $F$ direction is zero, only the velocity in the perpendicular $F$ direction $$ v ^ { \prime } = v \cos 37 ^ { \circ } = 4 \mathrm {~m} / \mathrm { s } $$ At this time, the direction of velocity is perpendicular to $F$ and the velocity is minimized, C is wrong; D. Perpendicular $F$ direction of the block to do uniform linear motion, there are $$ x _ { P Q } = v ^ { \prime } t ^ { \prime } = 12 \mathrm {~m} $$ D is correct.

Question 36

Question 36: 38. May 28, 2020, China's first domestically produced aircraft carrier "Shandong ship" in a sea area...

38. May 28, 2020, China's first domestically produced aircraft carrier "Shandong ship" in a sea area to carry out training missions. As shown in the figure assumes that a ship aircraft takeoff training, the mass of $m = 2 \times 10 ^ { 4 } \mathrm {~kg}$ aircraft in the catapult system under the action of $t _ { 1 } = 0.2 \mathrm {~s}$ to a certain initial speed into the deck runway, and after the deck to do a uniformly accelerated straight-line motion, after ${ } ^ { t _ { 2 } } = 4.0 \mathrm {~s}$ in the runway movement After ${ } ^ { t _ { 2 } } = 4.0 \mathrm {~s}$ on the runway, the airplane takes off after 120 m. The takeoff speed is $v = 50 \mathrm {~m} / \mathrm { s }$, not counting air resistance. The following statement is correct ![](/images/questions/phys-work-energy/image-027.jpg)

A. A. The magnitude of the momentum gained by the aircraft under the action of the ejection system is $1 \times 10 ^ { 5 } \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s }$
B. B. The average magnitude of the force acting on the aircraft by the ejection system is $1 \times 10 ^ { 6 } \mathrm {~N}$
C. C. The magnitude of the airplane's acceleration on the deck runway is $12.5 \mathrm {~m} / \mathrm { s } ^ { 2 }$
D. D. The work done on the airplane by the ejection system is $2.5 \times 10 ^ { 5 } \mathrm {~J}$

Answer

Answer: B

Solution: A. The initial velocity obtained by the airplane under the action of the catapult system is $v _ { 0 }$, the $$ \frac { v _ { 0 } + v } { 2 } t _ { 2 } = s $$ The solution is $$ v _ { 0 } = 10 \mathrm {~m} / \mathrm { s } $$ Then the magnitude of the momentum gained by the airplane under the action of the ejection system is $$ p _ { 0 } = m v _ { 0 } = 2 \times 10 ^ { 5 } \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } $$ Option A is wrong ; B. According to the momentum theorem $$ F t _ { 1 } = p _ { 0 } $$ The solution is $$ F = 1 \times 10 ^ { 6 } \mathrm {~N} $$ Option B is correct ; C. The magnitude of the acceleration of the airplane on the deck runway is $$ a = \frac { v - v _ { 0 } } { t _ { 2 } } = \frac { 50 - 10 } { 4 } \mathrm {~m} / \mathrm { s } ^ { 2 } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Option C is incorrect ; D.The work done by the ejection system on the airplane is $$ W = \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } = 1.0 \times 10 ^ { 6 } \mathrm {~J} $$ Option D is wrong.

Question 37

Question 37: 39. As shown in the figure, the mass are $m$ of the three balls from the height are $h$ of the top o...

39. As shown in the figure, the mass are $m$ of the three balls from the height are $h$ of the top of the smooth fixed inclined plane from the bottom of the bottom of the slide from a standstill, the three inclined plane inclination angle is different, the ![](/images/questions/phys-work-energy/image-028.jpg)

A. A. The work done by gravity on the ball is all ${ } ^ { m g h }$
B. B. The elastic force does work on the ball all ${ } ^ { m g h }$
C. C. The average power of the gravitational forces are all equal
D. D. The average power of the elastic forces are not equal

Answer

Answer: A

Solution: A. According to the characteristics of work done by gravity, the work done by gravity on the ball is ${ } ^ { m g h }$, option A is correct; B. The direction of the elastic force is perpendicular to the direction of displacement, then the elastic force does work on the ball are 0, option B is wrong; C. According to Newton's second law, the ball slides down the slope of the acceleration of gravity $a = g \sin \theta$, then the time of decline $$ t = \sqrt { \frac { 2 h } { a \sin \theta } } = \sqrt { \frac { 2 h } { g \sin ^ { 2 } \theta } } $$ Since the inclination angle $\theta$ of each inclined plane is different, the sliding time is different, and according to $\bar { P } _ { G } = \frac { W _ { G } } { t }$, we can know that the average power of gravity is not equal, and option C is wrong; D. According to $\overline { P _ { N } } = \frac { W _ { N } } { t }$, we can know that the average power of the elastic force is equal, option D is wrong;

Question 38

Question 38: 40. The carrier's carrier aircraft in the take-off process, only by its own engine jet is not enough...

40. The carrier's carrier aircraft in the take-off process, only by its own engine jet is not enough to reach the take-off speed on the flight deck, if the installation of auxiliary take-off electromagnetic catapult system (as shown in Figure A) will be able to reach the take-off speed. Electromagnetic catapult system of a design can be simplified as shown in Figure B, Figure $M N , P Q$ is a smooth parallel metal straight rail (resistance is negligible), $A B$ is the electromagnetic catapult car, loop $P B A M$ in the current is constant, the current produced by the magnetic field on the catapult car to exert The magnetic field generated by the current exerts a force on the catapult vehicle, which drives the carrier aircraft to accelerate to the right for takeoff from rest, regardless of air resistance, the following statements are correct about the system ![](/images/questions/phys-work-energy/image-029.jpg) Orbit A (side view) ![](/images/questions/phys-work-energy/image-030.jpg) B (top view) High School Physics Assignment, October 29, 2025

A. A. The magnetic field between $M N , P Q$ is a uniform magnetic field
B. B. The ejector car is accelerating with decreasing acceleration.
C. C. The kinetic energy of the ejected vehicle is proportional to the magnitude of the current.
D. D. Alternating current is applied to circuit PBAM and the ejector still accelerates normally.

Answer

Answer: D

Solution: A. According to the right-hand spiral law can be known between the parallel metal straight guide the existence of firm upward magnetic field, and the magnetic field generated by the energized straight wire for the toroidal magnetic field, the farther away from the wire, the weaker the magnetic field, the $M N , P Q$ between the field is not a uniformly strong magnetic field, A error; B. Along the guide direction, the magnetic field is unchanged, and $M , P$ ends loaded with a constant voltage, the resistance is unchanged, then the current size is also unchanged, the width of the parallel guide rail is also unchanged, by the amperometric force $$ F _ { \text {安 } } = B I L $$ It can be seen that the size of the ampere force is unchanged, so the acceleration of the ejected car is unchanged, so the ejected car to do uniformly accelerated straight-line motion, B error; C. According to the theorem of kinetic energy can be seen $$ E _ { \mathrm { k } } = F _ { \text {安 } } X $$ When the voltage increases, the current in the loop increases, and the current increases so that the magnetic field between the rails also increases, that is, the size of the voltage will affect the magnetic field and the size of the current, then by the $$ F _ { \text {安 } } = B I L $$ It can be seen that the amperometric force is not proportional to the magnitude of the voltage, so the kinetic energy is not proportional to the magnitude of the voltage, C error; D. According to the right-hand spiral law can be known along the current direction of the circuit PBAM, between the guide produces a firm upward magnetic field, combined with the left-hand rule can be known as electromagnetic catapult car by the direction of the amperage to the right; when the current direction along the circuit $M A B P$, according to the right-hand spiral law between the guide produces a firm downward magnetic field, combined with the left-hand rule can be known as electromagnetic catapult car by the direction of amperage is still to the right. The direction of the amperage force is still to the right. Therefore, the change of current does not change the direction of the amperometric force on the car, that is, the electromagnetic catapult system can work normally, D is correct.

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Work and Energy - Practice Questions (38)

Question 1: 1. As shown in the figure is a trampoline athlete performing in the air, in the process of the athle...

Question 2: 2. As shown in the figure, within the elastic limit, the spring with compression of $x$ is slowly st...

Question 3: 4. The diagram shows the effect of a stroboscopic photo of an apple falling freely. The apple is loc...

Question 4: 5. As shown in the figure, a student with a mass of 50 kg is doing sit-ups. If the mass of the stude...

Question 5: 6. The following physical quantities must change during the process of curved motion of the object i...

Question 6: 7. As shown in the figure, a mass of $m$ of the slider with an initial speed of ${ } ^ { V _ { 0 } }...

Question 7: 8. The small iron ball firm straight up and back to the throwing point in the process, then about th...

Question 8: 9. As shown in the figure, the ball is located on a smooth inclined plane, the inclined plane is loc...

Question 9: 10. An object moves in a straight line with variable speed. At ${ } ^ { t _ { 1 } }$ the rate is $v$...

Question 10: 11. As shown in the figure, a bullet with a horizontal speed into the original stationary block plac...

Question 11: 12. As shown in the figure, in the horizontal table on the $A$ point there is a mass of $m$ of the o...

Question 12: 13. As shown in the figure, a straight wooden board can be rotated in a straight plane around a hori...

Question 13: 14. Throw a ball with mass $m$ upward with a certain initial velocity, the maximum height of the bal...

Question 14: 15. As shown in the figure, the mass of the same two balls A, B, respectively, with a long $l$ and $...

Question 15: 16. In the derivation of "uniformly variable linear motion displacement formula", the entire process...

Question 16: 17. As shown in the figure, an object in 10 s along the horizontal road to the right of the movement...

Question 17: 18. As shown in the figure, with a line tethered ball in the smooth horizontal surface to do uniform...

Question 18: 19. As shown in the figure, the elevator mass is $M$, it is placed on the horizontal floor of a mass...

Question 19: 20. with a tension force F will be a weight of 10 N of the object raised at uniform speed 1 m, as sh...

Question 20: 22. As shown in the figure, Xiaoming pulls a wooden box with a light rope that makes an angle of $\t...

Question 21: A body of mass 2 kg is in free fall and lands in 2 s. The acceleration due to gravity is $\mathrm { ...

Question 22: 24. The following is a correct description of the understanding of work.

Question 23: When an object is thrown straight upwards, the object rises to a maximum height of $h$, and if the m...

Question 24: 26. A wooden block has a pulley at the front end, one end of the rope is tied to a fixed place on th...

Question 25: 27. A high-speed car due to brake failure, the driver will be the car into the outside of the ramp-t...

Question 26: 28. With regard to geostationary satellites, the following statements are incorrect ( ).

Question 27: 29. mass of $m$ objects along the bottom of the length are $L$, the angle of inclination is differen...

Question 28: 30. Manned flight bag is a single person low altitude flight device, as shown in the figure, its eng...

Question 29: 31. The following statements are correct

Question 30: 32. As shown in the figure for the investigation of the elastic potential energy of the spring with ...

Question 31: 33. An object of mass 1 kg slides on a horizontal surface, its kinetic energy varies with displaceme...

Question 32: 34. new coronavirus caused by the pneumonia epidemic touched the hearts of hundreds of millions of p...

Question 33: 35. December 11, 2020 Lianhuaiyang Town Railway line opened to traffic, realizing the dream of Yangz...

Question 34: 36. As shown in the figure, the length of $L$ light rod is fixed at one end of the mass of $m$ of th...

Question 35: 37. In the smooth horizontal plane, a slider mass $m = 2 \mathrm {~kg}$, in the horizontal plane by ...

Question 36: 38. May 28, 2020, China's first domestically produced aircraft carrier "Shandong ship" in a sea area...

Question 37: 39. As shown in the figure, the mass are $m$ of the three balls from the height are $h$ of the top o...

Question 38: 40. The carrier's carrier aircraft in the take-off process, only by its own engine jet is not enough...

Work and Energy

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Practicing topics ≠ Passing the exam

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