Assertion A: A 5 kg block on a rough horizontal table (coefficient of static friction 0.4, g = 10 m/s²) is pushed horizontally with 15 N and stays at rest, so the friction on it is 15 N. Reason R: Static friction adjusts itself to the applied force, up to its limiting value μₛN. Choose the correct answer.

Both A and R are correct and R is the correct explanation of A

Both A and R are correct but R is not the correct explanation of A

A is not correct but R is correct

A is correct but R is not correct

Laws of Motion

9 NEET-UG practice questions on Laws of Motion, part of Physics. The first 9 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Easy
What does the area under a force-time graph of a body represent?
1. APower
2. BImpulse
3. CAcceleration
4. DWork
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Answer: B · Impulse
Impulse is the integral of F dt, which is the area under the F-t graph, and it equals the change in momentum.
- A: Power is a rate, not an area under this graph.
- C: Acceleration is force divided by mass, not an area.
- D: Work is the area under a force-displacement graph.
Remember this
Area under F-t gives impulse and change in momentum; area under F-x gives work.
Question 2 · Hard
A 4 kg block on a rough horizontal table (coefficient of kinetic friction 0.25) carries a 2 kg block on top of it. A light string from the 4 kg block passes over a smooth pulley at the table's edge and supports a hanging 6 kg block. The system accelerates and the 2 kg block does not slip. Taking g = 10 m/s², find the friction force acting on the 2 kg block.
1. A7.5 N
2. B5 N
3. C10 N
4. D15 N
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Answer: A · 7.5 N
Friction from the table = 0.25 × 60 = 15 N, so a = (60 − 15)/12 = 3.75 m/s². The only horizontal force on the 2 kg block is friction, so f = 2 × 3.75 = 7.5 N. Check: T = 6(10 − 3.75) = 37.5 N, and 37.5 − 15 = 6 × 3.75.
- B: This applies μN with the table's coefficient to the upper block.
- C: This uses a = 5 m/s², which ignores friction at the table.
- D: This is the friction from the table on the lower block.
Remember this
A block carried without slipping feels only the friction needed for its acceleration, not μN.
Question 3 · Medium
Two blocks of 3 kg and 2 kg hang from the ends of a light string passing over a smooth, light pulley. Taking g = 10 m/s², what is the tension in the string once they are released?
1. A24 N
2. B20 N
3. C30 N
4. D25 N
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Answer: A · 24 N
a = (3 − 2)(10)/(3 + 2) = 2 m/s². For the lighter block, T = 2(10 + 2) = 24 N. Check with the heavier block: T = 3(10 − 2) = 24 N.
- B: This is the weight of the lighter block alone, ignoring its acceleration.
- C: This is the weight of the heavier block alone.
- D: This is the mean of the two weights, which is not what the equations of motion give.
Remember this
Tension lies between the two weights and equals 2m₁m₂g/(m₁ + m₂).
Question 4 · Medium
A curved road of radius 200 m is banked so that tan θ = 0.2. At what speed can a car take this curve with no help from friction? Take g = 10 m/s².
1. A28.3 m/s
2. B20 m/s
3. C44.7 m/s
4. D6.3 m/s
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Answer: B · 20 m/s
With no friction, the horizontal component of the normal force supplies the centripetal force, which gives tan θ = v²/(rg). So v² = 200 × 10 × 0.2 = 400 and v = 20 m/s.
- A: This uses v² = 2rg tan θ, an extra factor of 2.
- C: This uses √(rg), leaving out tan θ.
- D: This leaves out g, using √(r tan θ).
Remember this
The optimum banking speed is v = √(rg tan θ), independent of the mass of the car.
Question 5 · Medium
Assertion A: A 5 kg block on a rough horizontal table (coefficient of static friction 0.4, g = 10 m/s²) is pushed horizontally with 15 N and stays at rest, so the friction on it is 15 N. Reason R: Static friction adjusts itself to the applied force, up to its limiting value μₛN. Choose the correct answer.
1. ABoth A and R are correct but R is not the correct explanation of A
2. BA is not correct but R is correct
3. CBoth A and R are correct and R is the correct explanation of A
4. DA is correct but R is not correct
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Answer: C · Both A and R are correct and R is the correct explanation of A
The limiting friction is 0.4 × 50 = 20 N, more than the 15 N push, so the block stays at rest. Balance of forces makes the friction exactly 15 N, and this happens because static friction is self-adjusting. So A is true and R explains it.
- A: R is the very reason the friction takes the value 15 N.
- B: A is correct, since 15 N is below the limiting 20 N.
- D: R is a correct statement of static friction.
Remember this
Below the limit, static friction equals the applied force; μₛN is only its maximum value.
Question 6 · Medium
Friction is absent in the arrangement drawn, and the string and pulley are light. Once it is released, what tension does the string carry? Take g = 10 m/s².
Figure: MedExams
1. A20 N
2. B5 N
3. C30 N
4. D15 N
Answer and explanation in LearnFree account
Question 7 · Hard
Block A rests on block B, and the 42 N force acts on B alone. With g = 10 m/s² and equal static and kinetic friction coefficients, what are the accelerations of A and B?
Figure: MedExams
1. AA: 5 m/s², B: 10.5 m/s²
2. BA: 7 m/s², B: 7 m/s²
3. CA: 5 m/s², B: 8 m/s²
4. DA: 8 m/s², B: 5 m/s²
Answer and explanation in LearnFree account
Question 8 · Medium
Given below are two statements: Statement I: For a 2 kg block on a level surface with the friction graph shown (g = 10 m/s²), the coefficients of static and kinetic friction are 0.6 and 0.4. Statement II: Once the block in the graph slides, the friction force keeps growing as the applied force grows. Choose the correct answer.
Figure: MedExams
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer and explanation in LearnFree account
Question 9 · Easy
Net force on a 2 kg trolley, initially at rest on a straight track, varies with time as plotted. What is its speed at t = 4 s?
Figure: MedExams
1. A10 m/s
2. B40 m/s
3. C5 m/s
4. D20 m/s
Answer and explanation in LearnFree account

Laws of Motion

What does the area under a force-time graph of a body represent?

Two blocks of 3 kg and 2 kg hang from the ends of a light string passing over a smooth, light pulley. Taking g = 10 m/s², what is the tension in the string once they are released?

A curved road of radius 200 m is banked so that tan θ = 0.2. At what speed can a car take this curve with no help from friction? Take g = 10 m/s².

Friction is absent in the arrangement drawn, and the string and pulley are light. Once it is released, what tension does the string carry? Take g = 10 m/s².

Block A rests on block B, and the 42 N force acts on B alone. With g = 10 m/s² and equal static and kinetic friction coefficients, what are the accelerations of A and B?

Net force on a 2 kg trolley, initially at rest on a straight track, varies with time as plotted. What is its speed at t = 4 s?