Newton's Laws of Motion

Question 1

Question 1: 1. The three basic units of mechanics in the International System of Units are Read the followi...

1. The three basic units of mechanics in the International System of Units are Read the following material and complete the following 4 sub-questions, solving them without air resistance and taking the acceleration of gravity as $g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$. ![](/images/questions/phys-newton-laws/image-001.jpg) Xiaoming. ![](/images/questions/phys-newton-laws/image-002.jpg) 82 Follows 410 Fans 131 Weibo Tweets by Ming (131) ## March 15, 17:46 First time bungee jumping, shocked face! Exciting! Retweet (2845) Favorite Comment (1120) Standing on the 70-meter high tower, my legs are almost soft. Coach nimbly with spring rope buckle my feet. I was still muttering in my mind that this 30-meter rope is not long enough, it was gently pushed, "free fall journey" began! The earth came crashing down on me, $\cdots \cdots$, and when I was 10 meters short of having a close contact with the earth, the spring-loaded rope pulled me back into the air. Thrilling! Thrilling! Bungee jumping is the game of the brave, as for you believe it or not, I believe it anyway. ## Link Bungee jumping is a sport in which the body is bounced up and down in the air by jumping from a tower with a spring-loaded rope tied around the ankles that is fastened to the tower at one end.

A. A. N, kg, s
B. B. kg, m, s
C. C. N, m, s
D. D. N, kg, m

Answer

Answer: B

Solution: The three basic units of mechanics in the International System of Units are $\mathrm { kg } , \mathrm { m } , \mathrm { s }$. Therefore, choose B. ## 2. a 3. a 4. b 5. d [Knowledge Point] Known to be subjected to motion [Analyze]2.Ming swoops down to the ground and sees the earth coming down on him. So the motion of the earth is now described with Ming himself as the reference. So choose A. 3. rope length of 30 m, Ming in the drop 30 m, only by gravity, do free-fall motion, can be known to Ming free-fall motion displacement of 30 m. The length of the rope is 30 m, Ming in the drop 30 m, only by gravity, do free-fall motion. Therefore, choose A. 4. The first falling stage, the spring rope is stretched in the process, according to Newton's second law: $$ a = \frac { m g - F } { m } $$ The spring rope is stretched in the process, the elastic force becomes larger, then the acceleration becomes smaller, the direction downward, Xiaoming do acceleration decrease acceleration, when the elastic force is equal to the gravity, acceleration is zero, the maximum speed, thereafter the elastic force continues to increase, the acceleration direction upward, Xiaoming acceleration increase deceleration, so Xiaoming falling speed size change is the first to increase after the reduction of the situation, so the choice of the B . 5. rope length of 30 m, Ming in the fall of 30 m, only by gravity, do free-fall motion, so the spring force is equal to 0, so choose D.

Question 2

Question 2: 2. Ming feels "the earth pounding on his face", his system of reference is

2. Ming feels "the earth pounding on his face", his system of reference is

A. A. oneself
B. B. control tower
C. C. ground level
D. D. heavens

Answer

Answer: A

Question 3

Question 3: 3. Xiaoming first drop in the process of free-fall motion displacement of the size of about

3. Xiaoming first drop in the process of free-fall motion displacement of the size of about

A. A. 30 m
B. B. 40 m
C. C. 60 m
D. D. 70 m

Answer

Answer: A

Question 4

Question 4: 4. The first falling stage, the spring rope is stretched in the process, the change in the size of M...

4. The first falling stage, the spring rope is stretched in the process, the change in the size of Ming's falling speed is

A. A. It's getting bigger all the time.
B. B. increase and then decrease
C. C. decreasing then increasing
D. D. It's been decreasing.

Answer

Answer: B

Question 5

Question 5: 5. When Ming fell 30 meters, the size of the tension on the elastic rope is

5. When Ming fell 30 meters, the size of the tension on the elastic rope is

A. A. Equal to gravity
B. B. greater than the force of gravity
C. C. Less than gravity
D. D. equal to 0

Answer

Answer: D

Question 6

Question 6: 6. The following physical quantities belong to the basic physical quantities of the International Sy...

6. The following physical quantities belong to the basic physical quantities of the International System of Units.

A. A. current intensity
B. B. electric field strength
C. C. magnetic flux
D. D. magnetic field density

Answer

Answer: A

Solution: The International System of Units specifies seven fundamental physical quantities, namely, length, mass, time, thermodynamic temperature, electricity intensity of current, intensity of light, and quantity of matter.

Question 7

Question 7: 7. When a maglev train is running at high speed, the air resistance is proportional to the quadratic...

7. When a maglev train is running at high speed, the air resistance is proportional to the quadratic of its speed, i.e., $F _ { f } = k v ^ { 2 }$, and the unit of the proportionality coefficient $k$ is

A. A. kg/m
B. B. $\mathrm { kg } / \left( \mathrm { m } ^ { 2 } \mathrm {~s} ^ { 2 } \right)$
C. C. $\mathrm { N } / \left( \mathrm { m } ^ { 2 } \mathrm {~s} ^ { 2 } \right)$
D. D. $\mathrm { kg } / \mathrm { m } ^ { 3 }$

Answer

Answer: A

Solution: By $F _ { f } = k v ^ { 2 }$ $$ k = \frac { F _ { f } } { v ^ { 2 } } $$ The unit of force is N $$ 1 \mathrm {~N} = 1 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } ^ { 2 } $$ The units of $k$ can be obtained as $k g / m$.

Question 8

Question 8: 8. The following units are basic units in the International System of Units (SI)

8. The following units are basic units in the International System of Units (SI)

A. A. cow
B. B. hourly
C. C. meter (classifier)
D. D. Tibetan unit of land area, about 6 ares

Answer

Answer: C

Solution: A.The ox is a derived unit in the International System of Units, so A does not fit the question; BD . BD.The hour and the gram are not international units, so BD does not fit the question; C. The meter is the basic unit in the International System of Units, so C meets the meaning of the question.

Question 9

Question 9: 9. "Nature and the laws of nature were hidden in the night; and God said, Let Newton go. So everythi...

9. "Nature and the laws of nature were hidden in the night; and God said, Let Newton go. So everything was illuminated", nature in Newton and before like an open book, many of the mysteries of nature by Newton with mathematical methods to crack one by one. Newton as the founder of physics, for us to leave a lot of thought treasure. And which can be called the cornerstone of physics method is not only the later generation of physics people dissect the world scalpel is also students in each solution to the physics problem process must be done in the first step, this method refers to the

A. A. Constructing Physical Modeling Methods
B. B. deductive reasoning
C. C. reductio ad absurdum
D. D. controlled variable method

Answer

Answer: A

Solution: Constructing a physical modeling approach is a necessary first step in the process of solving a physics problem every time.

Question 10

Question 10: 10. The following units are basic units in the International System of Units (SIU)

10. The following units are basic units in the International System of Units (SIU)

A. A. unit of angle or arc equivalent one sixtieth of a degree
B. B. Sir Isaac Newton (1642-1727), British mathematician and physicist
C. C. watt (loan)
D. D. joule (loanword)

Answer

Answer: A

Solution: ABCD. There are only seven basic units in the International System of Units (SIU): the meter (m), the kilogram (kg), the second $( \mathrm { s } )$, the ampere (A), the kelvin (K), the mole (mol), and the candelabra (cd), and the rest are derived, so A is correct and BCD is wrong.

Question 11

Question 11: 11. The following units are basic units

11. The following units are basic units

A. A. Sir Isaac Newton (1642-1727), British mathematician and physicist
B. B. hourly
C. C. meter/second
D. D. meters/second ${ } ^ { 2 }$

Answer

Answer: B

Solution: "Hour" is the base unit of time; "newton", "meter/second" and "meter/second [[INLINE_FORMULA_0 ]] are derived units.

Question 12

Question 12: 13. With regard to the units of physical quantities, the following statements are correct (

13. With regard to the units of physical quantities, the following statements are correct (

A. A. Any physical quantity and physical concept has a corresponding unit
B. B. Physical quantities in physics formulas may also have no units, and such quantities have no numerical value
C. C. The unit relationship between physical quantities can not be determined without physical formulas that describe various laws
D. D. The units of physical quantities can all be derived from each other

Answer

Answer: C

Solution: Some physical quantities have no units, such as the dynamic friction factor, so A error; physical formula of physical quantities may not have units, but such quantities have numerical values, such as the dynamic friction factor, so B error; according to the unit system can be known, according to the physical formula to determine the relationship between the units of physical quantities, that is, physical quantities between the units of the relationship between the physical formulas that can not be separated from describing the laws of the physical formulas, so C is correct; basic physical quantities The units of the basic physical quantities are the basic units, can not be deduced from each other, so D is wrong.

Question 13

Question 13: 14. Some of the city's overpasses are equipped with intelligent electric escalators, as shown in the...

14. Some of the city's overpasses are equipped with intelligent electric escalators, as shown in the figure, when the passengers stand on the escalator, it first accelerates slowly, and then rises at a uniform speed, the... ![](/images/questions/phys-newton-laws/image-003.jpg)

A. A. Passengers are always overweight
B. B. Passengers are always weightless.
C. C. The force of the elevator on the passengers is always vertically upward
D. D. When the elevator rises at a constant speed, the force of the elevator on the passengers is firm and straight upward.

Answer

Answer: D

Solution: AB . Acceleration, the person has upward acceleration in an overweight state, uniform speed, is a state of equilibrium, AB error; C. Acceleration stage by the horizontal right friction, so the elevator on the person's force is not firm upward, C error; D. Elevator uniform speed, the combined external force is zero, that is, gravity and the elevator on the balance of the force on the person, D is correct.

Question 14

Question 14: 15. As shown in the figure, one of the following statements is true about a cart pulling a horse ver...

15. As shown in the figure, one of the following statements is true about a cart pulling a horse versus a horse pulling a cart

A. A. The horse pulling the cart doesn't move because the force of the horse pulling the cart is less than the force of the cart pulling the horse
B. B. The horse pulling the cart accelerates because the force of the horse pulling the cart is greater than the force of the cart pulling the horse
C. C. When a horse pulls a cart, the magnitude of the force of the horse pulling the cart is always equal to the magnitude of the force of the cart pulling the horse, regardless of whether the cart moves or does not move
D. D. When the horse pulls the cart without moving or the cart is moving at a constant speed, the force of the horse pulling the cart and the force of the cart pulling the horse are a pair of balanced forces

Answer

Answer: C

Question 15

Question 15: 16. As shown in the figure for a student found on the Internet a new energy vehicle cartoon, about t...

16. As shown in the figure for a student found on the Internet a new energy vehicle cartoon, about this cartoon, the following statements are correct ( ) ![](/images/questions/phys-newton-laws/image-004.jpg)

A. A. The force of the magnet on the iron block is greater than the force of the iron block on the magnet
B. B. The magnitude of the force of the magnet on the iron block is equal to the force of the iron block on the magnet
C. C. According to Newton's second law, this design causes the car to move forward with uniform acceleration
D. D. As long as the magnet is strong enough, the car can keep moving

Answer

Answer: B

Solution: AB. The force of the magnet on the iron block and the force of the iron block on the magnet is a pair of force and reaction, according to Newton's third law can be known, the force of the magnet on the iron block and the force of the iron block on the magnet are equal in size and opposite in direction, so A is wrong, B is correct; CD. Because the magnet and the iron block are part of the car, the force between them belongs to the internal force of the system, even if the magnet is strong enough, but for the car as a whole, the combined external force is zero, according to Newton's second law we can know that the acceleration is zero, the design can not make the car move forward, let alone make the car move all the time, so CD is wrong.

Question 16

Question 16: 17. As shown in the figure, the smooth horizontal surface is stationary with a car, alcohol lamp and...

17. As shown in the figure, the smooth horizontal surface is stationary with a car, alcohol lamp and test tube fixed in the car bracket constitutes the car as a whole, the alcohol lamp burns for a period of time after the plug spray. The following statement is correct ( ) ![](/images/questions/phys-newton-laws/image-005.jpg)

A. A. Since the mass of the plug is less than the mass of the cart, the impact force on the plug is greater than the impact force on the cart when it is ejected
B. B. Since the mass of the plug is less than the mass of the cart, the impact force on the plug is less than the impact force on the cart when it is ejected
C. C. At the moment the plug is ejected, the pressure of the cart on the horizontal surface is equal to the support of the cart on the horizontal surface
D. D. At the moment the plug is ejected, the pressure of the cart on the horizontal surface is less than the support force of the cart on the horizontal surface

Answer

Answer: C

Solution: AB . According to Newton's third law, the impact force on the plug and the impact force on the cart are equal in size and opposite in direction, so AB is wrong; CD. According to Newton's third law, the plug is sprayed out instantly, the pressure of the car on the horizontal surface is equal to the horizontal surface of the car's support, so C is correct D is wrong.

Question 17

Question 17: 18. The Shenzhou XVII spacecraft was launched successfully at 11:14 on October 26, 2023, in Jiuquan,...

18. The Shenzhou XVII spacecraft was launched successfully at 11:14 on October 26, 2023, in Jiuquan, and the crew of the mission consisted of three astronauts: Tang Hongbo, Tang Shengjie and Jiang Xinlin. As shown in the figure, it is the schematic diagram of measuring the mass of the space station by the kinetic method. If it is known that the mass of the spacecraft is $4 \times 10 ^ { 3 } \mathrm {~kg}$, and the average thrust of its propellers is 1280 N. After the spacecraft is docked to the space station, the change in the speed of the spacecraft and the space station is measured to be $0.07 \mathrm {~m} / \mathrm { s }$ within 7s of the work of the propellers. The following is correct ![](/images/questions/phys-newton-laws/image-006.jpg)

A. A. The acceleration of the spacecraft before docking with the space station is $0.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$
B. B. The acceleration of the spacecraft after docking with the space station is $0.1 \mathrm {~m} / \mathrm { s } ^ { 2 }$
C. C. The mass of the space station is $1.24 \times 10 ^ { 5 } \mathrm {~kg}$
D. D. The mass of the space station is $1.2 \times 10 ^ { 5 } \mathrm {~kg}$

Answer

Answer: C

Solution: A. According to Newton's second law, the acceleration of the spacecraft before docking with the space station is $$ a _ { 1 } = \frac { F } { m } = \frac { 1280 } { 4 \times 10 ^ { 3 } } \mathrm {~m} / \mathrm { s } ^ { 2 } = 0.32 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Therefore, A is wrong; B. The acceleration of the spacecraft after docking with the space station is $$ a _ { 2 } = \frac { \Delta v } { \Delta t } = \frac { 0.07 } { 7 } \mathrm {~m} / \mathrm { s } ^ { 2 } = 0.01 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Therefore, B is wrong; CD.Taking the spacecraft and the space station as a whole, according to Newton's second law, we can get $$ F = ( M + m ) a _ { 2 } $$ The mass of the space station is $$ M = 1.24 \times 10 ^ { 5 } \mathrm {~kg} $$ Therefore, C is correct and D is incorrect.

Question 18

Question 18: 19. Mobile phone to bring convenience to people at the same time also brought a lot of trouble, peop...

19. Mobile phone to bring convenience to people at the same time also brought a lot of trouble, people are more and more dependent on the cell phone, some people like to lie down to see the phone, often appear the phone smashed into the head of the situation. If the cell phone from the height of about 20 cm from the person without the initial speed of falling, hit the head of the cell phone did not rebound, the head of the impact of the cell phone time is about 0.1 s. Assuming that the force acting on the head of the cell phone is a constant force, the direction is straight down, take the acceleration of gravity $g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$, the following analysis is not correct ( ) ![](/images/questions/phys-newton-laws/image-007.jpg) ![](/images/questions/phys-newton-laws/image-008.jpg)

A. A. The magnitude of the velocity of the phone just before it is about to touch the head is approximately $2 \mathrm {~m} / \mathrm { s }$
B. B. The magnitude of the acceleration of deceleration of the cell phone when acting with the head is approximately $20 \mathrm {~m} / \mathrm { s } ^ { 2 }$
C. C. The force of the head on the phone causes the phone to generate an acceleration of approximately $20 \mathrm {~m} / \mathrm { s } ^ { 2 }$
D. D. The force of the cell phone on the head is approximately 3 times the force of gravity of the cell phone.

Answer

Answer: C

Solution: A. The cell phone is in free-fall motion just before it touches the head, then there are $$ v ^ { 2 } = 2 g h $$ where $$ h = 20 \mathrm {~cm} = 0.2 \mathrm {~m} $$ solves for $$ v = 2 \mathrm {~m} / \mathrm { s } $$ Therefore, A is correct; B. After the cell phone hits the head in uniformly decelerating motion, there is $$ a = \frac { v } { t } = \frac { 2 } { 0.1 } \mathrm {~m} / \mathrm { s } ^ { 2 } = 20 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Therefore, B is correct; CD. Let the force of the cell phone on the head be $F$ , by Newton's second law to get $$ F - m g = m a , a = 2 g $$ The force on the head is $F$. $$ F = 3 m g $$ The force of the head on the cell phone causes the cell phone to generate an acceleration of about the magnitude of $$ a ^ { \prime } = \frac { F } { m } = 3 g = 30 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Therefore, C is wrong D is correct ; The wrong choice for this question, so choose C.

Question 19

Question 19: 20. As shown in the figure, a horizontal conveyor belt $L = 10 \mathrm {~m}$ of length $A B$ is movi...

20. As shown in the figure, a horizontal conveyor belt $L = 10 \mathrm {~m}$ of length $A B$ is moving clockwise at a constant speed $v = 8 \mathrm {~m} / \mathrm { s }$ A block of mass $m = 2 \mathrm {~kg}$ slides onto the conveyor belt from the left end $v _ { 0 } = 4 \mathrm {~m} / \mathrm { s }$ of the conveyor belt at an initial speed $A$ and the kinetic friction factor between the block and the surface of the belt is $A$. _FORMULA_4]] slides onto a conveyor belt from the left end of the belt at $A$ The factor of kinetic friction between the block and the surface of the belt $\mu = 0.4 \left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$ is $A$. The following statement is correct ( ) ![](/images/questions/phys-newton-laws/image-009.jpg)

A. A. After 1 s, the work done by the sliding friction force on the block is - 48 J
B. B. After 1 s, the change in mechanical energy of the block is 64 J.
C. C. The kinetic energy of the block increases by 64 J as it travels from $A$ to $B$.
D. D. The internal energy of the system increases by 16 J as the block moves from $A$ to $B$.

Answer

Answer: D

Solution: AB. When the block is first placed on the conveyor belt, due to the relative motion with the conveyor belt, the block is subjected to sliding friction to the right, and the block is accelerated. $$ F _ { \mathrm { f } } = \mu m g = m a \Rightarrow a = 4 \mathrm {~m} / \mathrm { s } ^ { 2 } $$ After 1 s the displacement of the block relative to the ground is $$ x = v _ { 0 } t + \frac { 1 } { 2 } a t ^ { 2 } = 6 \mathrm {~m} < 10 \mathrm {~m} $$ The velocity of the block is $$ v = v _ { 0 } + a t = 8 \mathrm {~m} / \mathrm { s } $$ Sliding friction does positive work on the block $$ W = F _ { f } x = \mu m g x = 48 \mathrm {~J} $$ The change in mechanical energy of the block is 48 J . Therefore, AB is wrong; C. The block leaves the right end of the conveyor belt with the same speed as the conveyor belt. $$ \Delta E = \frac { 1 } { 2 } m v ^ { 2 } - \frac { 1 } { 2 } m v _ { 0 } ^ { 2 } = 48 \mathrm {~J} $$ D. Friction generates heat; D. Friction generates heat $$ Q = F _ { f } \cdot x _ { \text {相 } } = \mu m g ( v t - x ) = 0.4 \times 20 ( 8 \times 1 - 6 ) \mathrm { J } = 16 \mathrm {~J} $$ D. Friction generates heat $$ Q = F _ { f } \cdot x _ { \text {相 } } = \mu m g ( v t - x ) = 0.4 \times 20 ( 8 \times 1 - 6 ) \mathrm { J } = 16 \mathrm {~J} $$

Question 20

Question 20: 21. As shown in the figure, $a , b , c$ is a set of three blocks of mass $m$, which are $a , b$ conn...

21. As shown in the figure, $a , b , c$ is a set of three blocks of mass $m$, which are $a , b$ connected by a horizontal string and placed on a horizontal surface, and $c$ is placed $b$ on $a$. FORMULA_4]]. A horizontal tension is applied to $a$ so that the three objects move together horizontally to the right with uniform velocity. The kinetic friction factor between the contact surfaces is $\mu$ and the acceleration due to gravity is $g$. The following statement is correct ( ). ![](/images/questions/phys-newton-laws/image-010.jpg)

A. A. At this point the horizontal tension is equal to the elasticity of the light rope
B. B. The frictional force on the block $c$ is of the magnitude $\mu m g$
C. C. When this horizontal tension is increased to 1.5 times the original one, the frictional force on the block $c$ is $\mu \mathrm { mg }$
D. D. After cutting the light rope, during the movement of the block $b$ to the right, the magnitude of the friction force on the block $c$ is $\mu m g$

Answer

Answer: D

Solution: A. The three blocks move together in uniform linear motion, and by the equilibrium condition, for the $a , b , c$ system $$ F = 3 \mu m g $$ For the $b , c$ system $$ \mathrm { F } _ { \mathrm { T } } = 2 \mu m g $$ B. The $F > \mathrm { F } _ { \mathrm { T } }$ system $$ \mathrm { F } _ { \mathrm { T } } = 2 \mu m g $$ is a system of $F > \mathrm { F } _ { \mathrm { T } }$, i.e., the horizontal tension is greater than the elasticity of the rope; B. $c$ is in equilibrium with uniform linear motion, then $c$ is not subject to friction, so B is wrong; C.When the horizontal tension is increased to 1.5 times of the original [BLOCK_FORMULA_2]] is not subject to friction when the horizontal tension is increased by 1.5 times. By Newton's second law, the $a , b , c$ system is not subject to friction. $$ F ^ { \prime } - 3 \mu m g = 3 m a $$ For $c$ $$ f = m a $$ Solve for $$ f = 0.5 \mu \mathrm { mg } $$ Therefore, C is wrong; D. After cutting the light rope, during the movement of the block $b$ to the right, the acceleration of the block $b , c$ system is $$ a = \frac { 2 \mu m g } { 2 m } = \mu g $$ The amount of friction on the block $c$ is $$ f ^ { \prime } = m a ^ { \prime } = \mu m g $$ Therefore, D is correct.

Question 21

Question 21: 22. In the spring of 2024, the 50 kW-class double-ring nested Hall thruster developed by China Aeros...

22. In the spring of 2024, the 50 kW-class double-ring nested Hall thruster developed by China Aerospace Science and Technology Corporation (CASC) was successfully ignited and operated stably, marking China's ranking among the world's leaders in nested Hall electric propulsion technology. Nested Hall Thruster does not use traditional chemical propellant, but uses plasma propellant, which has a significant advantage of high specific impulse. Specific impulse is a physical quantity introduced by aerospace scientists to measure the fuel utilization efficiency of rocket engines, abbreviated as $I _ { \mathrm { sp } }$, which is the impulse generated by a unit mass of propellant, and the unit of this physical quantity should be ( ).

A. A. $\mathrm { m } / \mathrm { s }$
B. B. $\mathrm { kg } \cdot \mathrm { m } / \mathrm { s } ^ { 2 }$
C. C. $\mathrm { m } / \mathrm { s } ^ { 2 }$
D. D. N-s.

Answer

Answer: A

Solution: According to the meaning of the question than the impulse represents the impulse produced per unit mass of propellant, so it can be obtained that $$ I _ { \mathrm { sp } } = \frac { I } { m } $$ Combined with the momentum theorem $I = \Delta m v$, we can obtain the unit of the physical quantity of specific impulse as $$ 1 \frac { \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } } { \mathrm {~kg} } = 1 \mathrm {~m} / \mathrm { s } $$

Question 22

Question 22: 23 . The following statements about the mechanics unit system are false

23 . The following statements about the mechanics unit system are false

A. A. In analytical calculations in mechanics, only the International System of Units can be used, not other units.
B. B. The physical quantities chosen as base units in the mechanical unit system are length, time, mass
C. C. In the mechanical unit system, the basic units using the International System of Units (SI) are the kilogram, the meter, and the second.
D. D. Derived units in the system of units can be expressed in terms of base units

Answer

Answer: A

Solution: A. In analytical calculations related to mechanics, international units can be used, or other units. Therefore, A is wrong; B. In the unit system of mechanics, the physical quantities selected as the basic units are length, time and mass, so B is correct; C. In the mechanical unit system, the basic units of the International System of Units are kilograms, meters and seconds, so C is correct; D. The relationship between the physical quantities of the equation pushed out of the physical quantities of the unit is called the derived unit, the derived unit can be expressed in basic units, so D is correct. The question is incorrect, so choose A.

Question 23

Question 23: 24. On October 9, 2022, the Chinese Academy of Sciences observed the brightest gamma burst to date, ...

24. On October 9, 2022, the Chinese Academy of Sciences observed the brightest gamma burst to date, with an energy exceeding $10 ^ { 55 } \mathrm { erg }$. The correct expression for "erg" in terms of the basic units of the International System of Units is

A. A. $\mathrm { kW } \cdot \mathrm { h }$
B. B. J
C. C. $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } \cdot \mathrm {~s} ^ { - 2 }$
D. D. eV

Answer

Answer: C

Solution: Neither $\mathrm { ABD } . \mathrm { kW } \cdot \mathrm { h } , \mathrm {~J} , \mathrm { eV }$ is expressed in base units of the International System of Units (SI), and neither of the ABDs meets the meaning of the question; C. According to the definition of work $$ 1 \mathrm {~J} = 1 \mathrm {~N} \cdot \mathrm {~s} $$ According to Newton's second law we have $$ 1 \mathrm {~N} = 1 \mathrm {~kg} \cdot \mathrm {~m} / \mathrm { s } ^ { 2 } $$ Solve for $$ 1 \mathrm {~J} = 1 \mathrm {~kg} \cdot \mathrm {~m} ^ { 2 } \cdot \mathrm {~s} ^ { - 2 } $$

Question 24

Question 24: 25. A 45 kg person stands on a scale placed on the floor of a moving elevator and the scale reads 50...

25. A 45 kg person stands on a scale placed on the floor of a moving elevator and the scale reads 50 kg, then the motion of the elevator may be one of the following judgments ( (1) The lift accelerates upward (2) the lift accelerates down (3) The elevator decelerates to rise (4) The lift decelerates down

A. A. (1) (3)
B. B. (1) (4)
C. C. (2) (3)
D. D. (2) (4)

Answer

Answer: B

Solution: The mass of the person is 45 kg, when the scale shows 50 kg, it means that the support force on the person is greater than the force of gravity, according to Newton's second law, the acceleration of the person is upward; then the elevator may be accelerating, or decelerating. That is, (1) (4) two correct, (2) (3) two wrong. In summary, the answer to this question is B. [Eyes] overweight is the object by the limiting force, can also be called elastic force (tension or support) is greater than the object by the phenomenon of gravity. When the object to do upward acceleration or downward deceleration, the object are in the state of overweight, that is, regardless of how the object moves, as long as the acceleration is upward, the object is in the state of overweight.

Question 25

Question 25: 26. While riding in the train, Jun observes a dripping faucet in the train car. He notices that at o...

26. While riding in the train, Jun observes a dripping faucet in the train car. He notices that at one time the water drops fall at point $A$ to the right directly below the faucet, and at a later time the water drops fall at point $B$ directly below the faucet, as shown in the figure. Then the possible motion of the train during this process is ![](/images/questions/phys-newton-laws/image-011.jpg)

A. A. Decelerated motion to the left, then uniform motion
B. B. Uniform motion to the left, then decelerating motion
C. C. Decelerated motion to the right, then uniform motion
D. D. Accelerating motion to the left, then decelerating motion

Answer

Answer: C

Solution: When the water droplet falls, the horizontal direction keeps the original speed, if the car is stationary or moving at a constant speed, the horizontal displacement of the car is the same as that of the water droplet, and the water droplet will fall at $B$; if the car accelerates to the left, the water droplet will still keep the horizontal speed of the droplet, and the horizontal displacement of the car will increase, so the water droplet will fall at $A$; similarly, the car decelerates to the right, and the water droplet will fall at $A$. The water drop will fall at $A$; similarly, if the car decelerates to the right, the water drop will also fall at $A$. So choose C.

Question 26

Question 26: As shown in the figure, a light ring A on a thin straight rod is connected to a weight B by an inext...

As shown in the figure, a light ring A on a thin straight rod is connected to a weight B by an inextensible light rope across a fixed pulley, and an external force is applied so that A rises uniformly along the rod at a speed of $v$, and rises from the $M$ position in the figure to the point where the line with the fixed pulley is at the horizontal $N$ position. FORMULA_2]] position, it is known that $A O$ makes an angle of $\theta$ with the firm straight rod, then the following statement is wrong ( ) ![](/images/questions/phys-newton-laws/image-012.jpg)

A. A. At the beginning the speed of B is $v \cos \theta$
B. B. A Weight B decelerates as it rises uniformly
C. C. A moves to position $N$ when the velocity of B is maximum.
D. D. During the descent of weight B, the pull of the rope on B is greater than the gravitational force on B.

Answer

Answer: C

Solution: The velocity of B is equal to the partial velocity of A in the direction of the rope. [BLOCK_FORMULA_0]] Therefore, A is correct and does not satisfy the requirements of the question; BC. $$ v _ { \mathrm { B } } = v \cos \theta $$ A rises uniformly, $\theta$ increases, $\cos \theta$ decreases, and weight B decelerates; A moves to position $N$ when $\theta$ equals $90 ^ { \circ }$, and weight B moves to the lowest point, so the velocity of weight B is zero. ], weight B moves to the lowest point, the speed of weight B is zero, so B is correct and does not meet the requirements of the question, C is wrong and meets the requirements of the question; D. B falling process, do deceleration, acceleration direction up, the direction of the combined force up, the rope on the B tension is greater than the B gravity, so D is correct, does not meet the requirements of the question.

Question 27

Question 27: 29. High-pole boat skills is the cultural town of Jiaxing, Zhejiang Province (Wuzhen) is still prese...

29. High-pole boat skills is the cultural town of Jiaxing, Zhejiang Province (Wuzhen) is still preserved and perform traditional folk acrobatic arts, performers climbed on the bamboo poles fixed in the boat, simulating silkworms spitting cocoon action to pray for cocoon harvest. As shown in the figure, at this time the performer is stationary in the curved tilt of the bamboo pole, then the following statement is correct ( ) ![](/images/questions/phys-newton-laws/image-013.jpg)

A. A. The performer's elastic force on the bamboo pole is produced by the deformation of the pole
B. B. The performer's force on the pole is directed straight downward.
C. C. The friction of the performer on the pole must be zero.
D. D. The force of the performer on the pole is greater than the force of the pole on the performer.

Answer

Answer: B

Solution: The elastic force of the performer on the bamboo pole is generated by the deformation of the person, so A is wrong; because the performer is stationary on the curved and tilted bamboo pole, gravity, elasticity and friction, and is in equilibrium, so the force of the bamboo pole on the performer is firm and straight upward, and according to Newton's third law can be known to the performer on the bamboo pole is firm and straight downward, so B is correct and C is wrong; according to Newton's third law can be known to the performer on the bamboo pole is equal to the force of the bamboo pole, so D is wrong; according to Newton's third law can be known to the performer on the bamboo pole, so D is wrong. According to Newton's third law, the force of the performer on the pole is equal to the force of the performer on the pole, so D is wrong. The force of the performer on the pole is equal to the force of the pole on the performer, so D is wrong. [INLINE_FORMULA_0]] [Knowledge Points] The use of equilibrium inference to find the magnitude or direction of the force, Newton's third law [Detailed explanation] when carrying a spring dynamometer will be submerged in the water, the iron ball due to upward buoyancy, the spring dynamometer shows a decrease; at the same time due to the ball of the water to produce downward reaction force, so that the scales show a larger number.

Question 28

Question 28: 30. A physics interest group with the device shown in the figure to study the interaction between ob...

30. A physics interest group with the device shown in the figure to study the interaction between objects. A cup of water on a scale, a spring loaded dynamometer attached to an iron ball, respectively, read the scale and the spring loaded dynamometer. Then lift the spring dynamometer will be submerged in water, then ( ) ![](/images/questions/phys-newton-laws/image-014.jpg) ![](/images/questions/phys-newton-laws/image-015.jpg)

A. A. The scale remains unchanged.
B. B. Decrease in the number of weighing scales
C. C. The spring scale decreases
D. D. Increase in the spring scale

Answer

Answer: C

Question 29

Question 29: 31. Many physical quantities in the natural sciences have more than one expression, usually with a d...

31. Many physical quantities in the natural sciences have more than one expression, usually with a definitional and a deterministic form, which reflect different levels of understanding of the natural world. Definitions focus on describing the objective world, while determinants focus on explaining cause-and-effect relationships. The following expressions focus on the explanation of cause and effect.

A. A. $a = \frac { F } { m }$
B. B. $R = \frac { U } { I }$
C. C. $\varphi = \frac { E _ { \mathrm { p } } } { q }$
D. D. $E = \frac { F } { q }$

Answer

Answer: A

Solution: A. Acceleration occurs because of force, so acceleration $a = \frac { F } { m }$ focuses on causation, so A is correct; B. $R = \frac { U } { I }$ is the defining equation of resistance, not the determining equation, and does not focus on causation, so B is wrong; C. $\varphi = \frac { E _ { \mathrm { p } } } { q }$ is the defining equation of potential, not the determining equation, and it does not focus on the causal relationship, so C is wrong; D. $E = \frac { F } { q }$ is the defining formula of electric field strength, not the determining formula, and it does not focus on explaining the causal relationship, so D is wrong.

Question 30

Question 30: 32 . The following statements are correct

32 . The following statements are correct

A. A. Galileo believed that the heavier the object, the faster it fell.
B. B. Aristotle believed that force is not what maintains the motion of an object, but what changes its state of motion
C. C. Newton formulated the law of gravity and measured the gravitational constant
D. D. Newtonian mechanics can solve the problem of low velocity motion of macroscopic objects

Answer

Answer: D

Solution: A. Aristotle believed that the heavier an object is, the faster it falls, so A is wrong; B.Galileo believed that force is not the cause of maintaining the state of motion of an object, and Newton's first law states that force is not the cause of maintaining the state of motion of an object, but force is the cause of changing the state of motion of an object, so B is wrong; C. The law of gravity was proposed by Newton, but the gravitational constant was measured by Cavendish, so C is wrong; D. Newtonian mechanics can solve the problem of low-speed motion of macroscopic objects, so D is correct.

Question 31

Question 31: 33. As shown in the figure, the smooth horizontal surface placed on the mass were $m , 3 m$ of $\mat...

33. As shown in the figure, the smooth horizontal surface placed on the mass were $m , 3 m$ of $\mathrm { A } , \mathrm {~B}$ two objects, $\mathrm { A } , \mathrm {~B}$ the maximum static friction between $\mu \mathrm { mg }$, is now a horizontal tension [$F$ pull B, so that A, B move with the same acceleration, then the maximum value of the tension $F$ is $F$. INLINE_FORMULA_4]] is used to pull B, so that A and B move with the same acceleration, then the maximum value of the tension $F$ is ![](/images/questions/phys-newton-laws/image-016.jpg)

A. A. $4 \mu m g$
B. B. $3 \mu m g$
C. C. $2 \mu m g$
D. D. $\mu m g$

Answer

Answer: A

Solution: When the static friction between $\mathrm { A } , \mathrm {~B}$ reaches its maximum, the tension $F$ reaches its maximum, and for object A, according to Newton's second law, there are $$ \mu m g = m a $$ The solution is $$ a = \mu g $$ For A and B as a whole, according to Newton's second law, we have $$ F = ( 3 m + m ) a $$ The association gives $$ F = 4 \mu \mathrm { mg } $$

Question 32

Question 32: 34. One of the correct statements about the following forces is that

34. One of the correct statements about the following forces is that

A. A. An object subjected to friction must be subjected to an elastic force, and the two forces must be perpendicular to each other in direction
B. B. The direction of friction on a moving object must be opposite to the direction of its motion.
C. C. The magnitude of the same pair of action and reaction may sometimes be unequal.
D. D. An object in total weightlessness is not subject to gravity

Answer

Answer: A

Solution: A. According to the conditions of friction, it can be seen that the friction force must be subjected to elastic force; the direction of the elastic force is perpendicular to the contact surface. The direction of friction is tangent to the contact surface, so the direction of these two forces must be perpendicular to each other, so A is correct; B. The direction of the friction force on the moving object must be opposite to the direction of its relative motion, when the friction force is power, the direction of the friction force is the same as the direction of motion, so B is wrong; C. The same pair of force and reaction are equal in size, so C is wrong; D. completely weightless state of the object only by gravity, so D error;

Question 33

Question 33: 35. As shown in Figure A, the mass of $m$ of the small rocket accelerated from rest, the acceleratio...

35. As shown in Figure A, the mass of $m$ of the small rocket accelerated from rest, the acceleration $a$ and the inverse of the speed of $\frac { 1 } { v }$ of the relationship between the image shown in Figure B, the rocket's speed is $v _ { 1 }$, the corresponding acceleration is $a _ { 1 }$. The acceleration at $a _ { 1 }$ is $a _ { 1 }$, and the gravitational acceleration is $g$ without counting the air resistance and the mass loss of the fuel burning. ![](/images/questions/phys-newton-laws/image-017.jpg) A ![](/images/questions/phys-newton-laws/image-018.jpg) B

A. A. The rocket starts at constant power.
B. B. The power of the rocket is $m \left( a _ { 1 } + g \right) V _ { 1 }$
C. C. $a - \frac { 1 } { v }$ The slope of the relational image is $\frac { a _ { 1 } + g } { v _ { 1 } }$
D. D. $a - \frac { 1 } { v }$ The intercept of the horizontal axis of the relational image is $\frac { g } { \left( a _ { 1 } + g \right) v _ { 1 } }$

Answer

Answer: C

Solution: According to the question, let the power of the rocket be $P$, the traction force be $F$, and from the formula $P = F v$ we get $$ F = \frac { P } { v } $$ ABC. According to the meaning of the question, by Newton's second law have $$ F - m g = m a $$ The solution is $$ a = \frac { P } { m } \cdot \frac { 1 } { v } - g $$ From Figure B, the slope of the image of the $a - \frac { 1 } { v }$ relation is $$ k = \frac { a _ { 1 } - ( - g ) } { \frac { 1 } { v _ { 1 } } } = \left( a _ { 1 } + g \right) v _ { 1 } $$ Then we have $$ \frac { P } { m } = \left( a _ { 1 } + g \right) v _ { 1 } $$ can be obtained $$ P = m \left( a _ { 1 } + g \right) V _ { V _ { 1 } } $$ That is, the rocket is started with constant power, so AB is correct and does not meet the meaning of the question; C is wrong and meets the meaning of the question; D. From the above analysis, it can be seen that the relational equation of $a - \frac { 1 } { v }$ is $$ a = \frac { P } { m } \cdot \frac { 1 } { v } - g $$ When $a = 0$, the solution is $$ \frac { 1 } { v } = \frac { m g } { P } = \frac { g } { \left( a _ { 1 } + g \right) v _ { 1 } } $$ That is, the intercept of the horizontal axis of the relational image of $a - \frac { 1 } { v }$ is $\frac { g } { \left( a _ { 1 } + g \right) v _ { 1 } }$, so D is correct and does not meet the meaning of the question.

Question 34

Question 34: 36. As shown in the figure, a soccer ball with three columns (can be regarded as a mass) support, sm...

36. As shown in the figure, a soccer ball with three columns (can be regarded as a mass) support, small columns into a positive triangle on the horizontal ground, the columns and the center of the soccer ball line with the direction of the firm straight angle are $30 ^ { \circ }$, known as the mass of the soccer ball $m$, the acceleration of gravity is $g$, not counting the friction between the soccer ball and the soccer ball, then the pressure on each column by the soccer ball is ( ). FORMULA_2]], not counting the friction between the soccer ball and the columns, then the pressure of each column by the soccer ball is ( ) ![](/images/questions/phys-newton-laws/image-019.jpg) Side View ![](/images/questions/phys-newton-laws/image-020.jpg) Top view

A. A. $\frac { 1 } { 6 } m g$
B. B. $\frac { 2 } { 3 } m g$
C. C. $\frac { 2 \sqrt { 3 } } { 9 } m g$
D. D. $\frac { 1 } { 3 } m g$

Answer

Answer: C

Solution: The forces on the soccer ball are analyzed as follows, with each post supporting the soccer ball $N$ diagonally upward and clamped $30 ^ { \circ }$ to the firm and straight direction. ![](/images/questions/phys-newton-laws/image-021.jpg) By decomposing all three $N$ orthogonally to the horizontal and straight directions, the equilibrium condition of the co-point force has $$ 3 N \cos 30 ^ { \circ } = m g $$ The solution is given by $$ 3 N \cos 30 ^ { \circ } = m g $$ $$ N = \frac { 2 \sqrt { 3 } } { 9 } m g $$ By Newton's third law, the pressure on each post from the soccer ball is $\frac { 2 \sqrt { 3 } } { 9 } m g$ in the direction of $30 ^ { \circ }$, and in the direction of $30 ^ { \circ }$. The direction is $30 ^ { \circ }$ and the direction is $30 ^ { \circ }$.

Question 35

Question 35: 37 . During a curling match, an athlete may use a curling brush to brush the ice in the path of the ...

37 . During a curling match, an athlete may use a curling brush to brush the ice in the path of the curling to reduce the kinetic friction factor between the curling and the ice. As shown in the figure, during a curling match, a curling ball (which can be regarded as a mass point) travels at a speed of $v$ from $A$ point vertically into the three rectangular area along the dotted line to do deceleration, arrived at $B$ at the rate of 0.5 v, with the curling brush with different strength to brush the $B C , C D$ between the surface of the ice, the curling movement to $C$ at the rate of 0.5 v, with different strength brush $C$, the curling movement to the $0.2 v$ at the rate of 0.4 v. 4]] with a velocity of $0.2 v$ and zero when it is just about to leave the $D$ point at the edge of the third rectangular area, and it is known that the kinetic friction factor of the curler on the ice between $A B , B C , C D$ is $A B , B C , C D$, respectively, $\mu _ { 1 } , \mu _ { 2 } , \mu _ { 3 }$. $\mu _ { 1 } , \mu _ { 2 } , \mu _ { 3 }$ respectively, and the movement time of the curling on the ice between $A B , B C , C D$ is $t _ { 1 } , t _ { 2 } , t _ { 3 }$ respectively, then ( ) ![](/images/questions/phys-newton-laws/image-022.jpg)

A. A. $\mu _ { 1 } : \mu _ { 2 } : \mu _ { 3 } = 5 : 3 : 2$
B. B. $\mu _ { 1 } : \mu _ { 2 } : \mu _ { 3 } = 25 : 9 : 4$
C. C. $t _ { 1 } : t _ { 2 } : t _ { 3 } = 14 : 30 : 105$
D. D. $t _ { 1 } : t _ { 2 } : t _ { 3 } = ( \sqrt { 3 } - \sqrt { 2 } ) : ( \sqrt { 2 } - 1 ) : 1$

Answer

Answer: C

Question 36

Question 36: 38. As shown in the figure, there is a box full of potatoes, with a certain initial speed in the kin...

38. As shown in the figure, there is a box full of potatoes, with a certain initial speed in the kinetic friction factor of $\mu$ of the horizontal ground sliding to do uniform deceleration, known as potatoes together with the box of the total mass of $M$, not counting the other external forces and air resistance, one of the mass of of the potato $A$ is stationary with respect to the box, then $A$ is subject to the force $F$ exerted by the other potatoes on it, and the size of the force $F = \mu m g$ is ( ). ![](/images/questions/phys-newton-laws/image-023.jpg)

A. A. $F = \mu m g$
B. B. $F = m g$
C. C. $F = \sqrt { ( m g ) ^ { 2 } + ( \mu M g ) ^ { 2 } }$
D. D. $F = m g \sqrt { 1 + \mu ^ { 2 } }$

Answer

Answer: D

Solution: Analyzing the potato, the potato is subject to the force of gravity, the force of other potatoes on it ![](/images/questions/phys-newton-laws/image-024.jpg) According to Newton's second law we have $$ F _ { \text {合 } } = M a = \mu M g $$ Solve for $$ a = \mu g $$ According to the parallelogram rule, the force of the other potatoes on it $$ F = \sqrt { ( m g ) ^ { 2 } + ( \mu m g ) ^ { 2 } } = m g \sqrt { 1 + \mu ^ { 2 } } $$ ABC is incorrect and D is correct.

Question 37

Question 37: 39. Drones are widely used in many fields such as aerial photography, rescue and disaster relief, an...

39. Drones are widely used in many fields such as aerial photography, rescue and disaster relief, and environmental monitoring. As shown in the figure, the mass of $M$ is $M$. As shown in the figure, a UAV of mass $M$ hangs a medical package of mass $m$ along the horizontal direction to do a uniformly accelerated straight-line motion, the angle between the light rope and the direction of the straight $\theta = 60 ^ { \circ }$, and the gravitational acceleration is $g$, regardless of the force of the parcels by the air. The following statement is correct () ![](/images/questions/phys-newton-laws/image-025.jpg) $( M + m ) g$

A. A. The drone's acceleration magnitude is $\sqrt { 3 } g$
B. B. The magnitude of the force of air on the drone is
C. C. The magnitude of the tension in the light rope is 2 Mg
D. D. If the light rope suddenly breaks, the package will be in free fall motion

Answer

Answer: A

Solution: A. The force analysis of the parcel shows that $$ m g \tan 60 ^ { \circ } = m a $$ The magnitude of the acceleration of the drone is solved as $$ a = \sqrt { 3 } g $$ Option A is correct; B. Consider the drone and the package as a whole, the whole is only subject to gravity and air force, the combined direction of the force is horizontally to the right, according to the rule of triangles can be known that the drone is subjected to the size of the air force is $$ F = ( M + m ) \sqrt { g ^ { 2 } + a ^ { 2 } } $$ Solution $$ F = 2 ( M + m ) g $$ Therefore, B is wrong; C. The parcel is balanced by the force in the firm direction, according to the equilibrium conditions there are $$ T \cos 60 ^ { \circ } = m g $$ So the tension on the rope is $$ T = 2 m g $$ Therefore, C is wrong; D. If the light rope suddenly breaks, the package will be thrown horizontally because it has horizontal velocity, so D is wrong. D. The package will be thrown horizontally, so D is wrong. $40 . \mathrm { B }$ [Knowledge Points] to determine whether there is static friction and its direction, the size and direction of sliding friction, Newton's third law [Detailed Explanation]AB. When the direction of the speed of the brush is horizontally to the right, the friction of the red paper on the brush is to the left, according to the interaction force, the friction of the brush on the red paper is to the right, according to the balance of forces, the friction of the desktop on the red paper is to the left. The friction of the red paper on the desktop to the right, so A is wrong, B is correct; CD. to paperweight as an object, according to the balance of forces can be known, the red paper has no friction on the paperweight, the paperweight also has no friction on the red paper, so C D error.

Question 38

Question 38: 40 . Chinese New Year Festival, Xiao Qiang put the red paper spread on the horizontal desktop to wri...

40 . Chinese New Year Festival, Xiao Qiang put the red paper spread on the horizontal desktop to write the word "Fu", as shown in the figure, in order to prevent the red paper from sliding, put a paperweight on its left to press, when the direction of the speed of the brush horizontally to the right, the following statements are correct () ![](/images/questions/phys-newton-laws/image-026.jpg)

A. A. The friction of the red paper on the brush is horizontal to the right.
B. B. The friction of the red paper on the tabletop is horizontal to the right
C. C. The friction of the paperweight on the red paper is horizontal to the right
D. D. The friction of the red paper against the paperweight is horizontal and to the right.

Answer

Answer: B

Question 39

Question 39: 41 . As shown in the figure, many subway stations have installed intelligent step escalators. In ord...

41 . As shown in the figure, many subway stations have installed intelligent step escalators. In order to save energy, when there are no passengers, that is, when the escalator is not loaded tend to run at a small speed, when there are passengers when the escalator after the first acceleration and then the uniform speed of two stages of movement. A passenger by the step escalator up the stairs, it happens to have gone through these two processes. The following statements are correct ( ) ![](/images/questions/phys-newton-laws/image-027.jpg)

A. A. As the escalator accelerates, the friction of the escalator on the customer is in the horizontal direction
B. B. When the escalator accelerates and runs at a constant speed, the direction of friction on the customer is horizontal.
C. C. The magnitude of the bouncing force of the escalator on the customer is always equal to the force of gravity
D. D. Customers are always subject to three forces

Answer

Answer: A

Solution: Passengers in the process of acceleration up, the combined force along the elevator diagonally upward, at this time the escalator on the customer's elastic force is greater than the force of gravity, due to the direction of friction and the direction of the relative movement of the opposite direction, so the friction force is along the horizontal direction; when the passengers rise at a constant rate of speed in the course of the escalator on the customer's elastic force and the force of gravity is equal in size and opposite in direction, friction force is equal to zero, at this time the passenger is in equilibrium, the effect of the two forces.

Question 40

Question 40: 42. As shown in the figure, a skier flies out from $M$ on a mountain with a horizontal velocity $v _...

42. As shown in the figure, a skier flies out from $M$ on a mountain with a horizontal velocity $v _ { 0 }$, and lands on the slope at $N$ after $t _ { 0 }$ for a period of time, when the direction of the velocity is just down along the slope, and then slides freely down a straight line $N P$ from point $N$. INLINE_FORMULA_4]] point, then from $N$ point along the straight line $N P$ free sliding, and after $t _ { 0 }$ time to reach the bottom of the slope at $P$ point. The angle between the slope $N P$ and the horizontal plane is $30 ^ { \circ }$, and frictional resistance and air resistance are not taken into account, then the acceleration of the skier and the change of velocity with time from $M$ to $P$ are correctly modeled as follows The correct image of the change in acceleration and velocity over time is ![](/images/questions/phys-newton-laws/image-028.jpg)

A. A. ![](/images/questions/phys-newton-laws/image-001.jpg)
B. B. ![](/images/questions/phys-newton-laws/image-002.jpg)
C. C. ![](/images/questions/phys-newton-laws/image-003.jpg)
D. D. ![](/images/questions/phys-newton-laws/image-004.jpg)

Answer

Answer: A

Solution: During $\mathrm { AB } . M$ to $N$, the skier makes a parabolic motion with an acceleration of $g$, which remains unchanged, and then enters the slope and makes a uniformly accelerated straight-line motion with an acceleration of $$ a = \frac { m g \sin 30 ^ { \circ } } { m } = g \sin 30 ^ { \circ } = \frac { 1 } { 2 } g $$ B is wrong A is correct; CD.The velocity during the parabolic motion $$ v = \sqrt { v _ { 0 } ^ { 2 } + g ^ { 2 } t ^ { 2 } } $$ It can be seen that $v$ is not linearly related to $t$, and after entering the slope, it is in uniformly accelerated straight-line motion, and its speed increases uniformly with time.

Question 41

Question 41: 43. As shown in Figure A, the mass of the spring is not counted vertically fixed in the horizontal p...

43. As shown in Figure A, the mass of the spring is not counted vertically fixed in the horizontal plane, $t = 0$ time, will be a metal ball from the spring is above a certain height by the static release, the ball fell to the spring on the spring compression spring to the lowest point, and then was bouncing away from the spring, rise to a certain height and then fall, and so on and so forth. Installed in the lower end of the spring pressure sensor, measured this process spring force $F$ with the time $t$ change in the graph shown in Figure B, then ( ) ![](/images/questions/phys-newton-laws/image-029.jpg) ![](/images/questions/phys-newton-laws/image-030.jpg) (a) ![](/images/questions/phys-newton-laws/image-031.jpg) (B) High School Physics Assignment, October 29, 2025

A. A. $t _ { 1 }$ Moment of maximum kinetic energy of the ball.
B. B. $t _ { 2 }$ Moment of maximum kinetic energy of the ball.
C. C. $t _ { 2 } \sim t _ { 3 }$ During this time, the mechanical energy of the ball and spring system is conserved
D. D. $t _ { 2 } \sim t _ { 3 }$ During this time, the sum of the ball's kinetic energy and gravitational potential energy is decreasing

Answer

Answer: C

Solution: A. ${ } ^ { t _ { 1 } }$ moment the ball just contact with the spring, the ball's gravity is greater than the spring force, still accelerating, the speed is not the maximum; when the spring force and gravity balance the speed is maximum, so A error; B. ${ } ^ { t }$ moment, the elastic force $F$ is the largest, so the spring compression is the largest, the ball movement to the lowest point, the speed is equal to 0, the kinetic energy is 0, so B is wrong; CD. ${ } ^ { t _ { 2 } } \sim t _ { 3 }$ time, the ball and the spring system mechanical energy conservation, ${ } ^ { t _ { 3 } }$ moment ball and spring detachment, so the ball increased kinetic energy and gravitational potential energy and equal to the spring to reduce the potential energy of the elasticity of the spring, so C is correct, D is wrong.

Newton's Laws of Motion

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