Question 1

A force of 20 N acts on a mass of 4 kg. What is the acceleration produced?

Accepted Answer

Using Newton's second law, F = ma, so a = F/m = 20/4 = 5 m/s².

Question 2

What is the SI unit of force according to Newton's second law of motion?

Accepted Answer

The SI unit of force is Newton (N), defined as 1 kg·m/s² from F = ma.

Question 3

A 2 kg object accelerates at 3 m/s². What net force acts on it?

Accepted Answer

Using F = ma, F = 2 × 3 = 6 N.

Question 4

According to Newton's third law of motion, if object A exerts a force on object B, then:

Accepted Answer

Newton's third law states that forces always occur in equal and opposite pairs between two interacting objects.

Question 5

A constant force of 50 N is applied to a 10 kg box on a frictionless surface. If the box starts from rest, what distance does it travel in 4 seconds?

Accepted Answer

First find acceleration: a = F/m = 50/10 = 5 m/s²; then use s = ½at² = ½(5)(4²) = 40 m.

Question 6

A force of 20 N is applied on a body of mass 4 kg. If the body was initially at rest, what is its acceleration?

Accepted Answer

Using Newton's second law: F = ma, so a = F/m = 20/4 = 5 m/s².

Question 7

A body of mass 2 kg moves with a velocity of 10 m/s. What force is required to stop it in 5 seconds?

Accepted Answer

Acceleration = (v - u)/t = (0 - 10)/5 = -2 m/s²; Force = m × a = 2 × 2 = 4 N.

Question 8

According to Newton's first law of motion, an object at rest will remain at rest unless acted upon by an external force. What is this property called?

Accepted Answer

The tendency of an object to resist changes in its state of motion is called inertia, which is Newton's first law.

Question 9

Two blocks of masses 3 kg and 5 kg are in contact on a frictionless surface. A force of 40 N is applied to the 3 kg block. What is the acceleration of the system?

Accepted Answer

Total mass = 3 + 5 = 8 kg; acceleration = F/(total mass) = 40/8 = 5 m/s².

Question 10

A car of mass 1000 kg experiences a friction force of 500 N while moving. If an engine force of 2000 N is applied, what is the net acceleration?

Accepted Answer

Net force = 2000 - 500 = 1500 N; acceleration = 1500/1000 = 1.5 m/s².

Question 11

Which of the following statements correctly defines Newton's third law of motion?

Accepted Answer

Newton's third law states that forces always occur in pairs: if object A exerts a force on B, then B exerts an equal and opposite force on A.

Question 12

A 6 kg block is pulled by a rope with a tension of 48 N on a frictionless horizontal surface. What distance does it travel in 4 seconds starting from rest?

Accepted Answer

Acceleration = 48/6 = 8 m/s²; distance = ½at² = ½ × 8 × 16 = 64 m.

Question 13

A ball of mass 0.5 kg is thrown upward with an initial velocity of 20 m/s. Considering g = 10 m/s², what is the net force acting on the ball during its upward motion?

Accepted Answer

The only force acting is gravity; F = mg = 0.5 × 10 = 5 N downward, regardless of the ball's velocity.

Question 14

A truck of mass 500 kg accelerates uniformly from 0 to 30 m/s in 10 seconds on a straight road. What is the magnitude of the force applied by the engine (neglecting friction)?

Accepted Answer

Acceleration = (30 - 0)/10 = 3 m/s²; Force = ma = 500 × 3 = 1500 N.

Question 15

A block of mass 5 kg is pushed along a frictionless horizontal surface with a force of 20 N. If the block starts from rest, what is its acceleration?

Accepted Answer

Using Newton's second law, F = ma, so a = F/m = 20/5 = 4 m/s².

Question 16

Two blocks A (mass 3 kg) and B (mass 2 kg) are connected by a light string on a frictionless surface. A horizontal force of 15 N is applied to block A. What is the tension in the string connecting them?

Accepted Answer

Total acceleration = 15/(3+2) = 3 m/s²; tension in string = force needed to accelerate block B = 2 × 3 = 6 N.

Question 17

A car of mass 1000 kg moving at 20 m/s brakes with a constant deceleration and comes to rest in 5 seconds. What is the magnitude of the braking force applied?

Accepted Answer

Deceleration a = (v - u)/t = (0 - 20)/5 = -4 m/s²; braking force F = m|a| = 1000 × 4 = 4000 N.

Question 18

A body of mass 2 kg experiences two perpendicular forces of 6 N and 8 N. According to Newton's second law, what is the resultant acceleration of the body?

Accepted Answer

Resultant force = √(6² + 8²) = √(36 + 64) = 10 N; acceleration a = F/m = 10/2 = 5 m/s².

Agniveer Army Technical Laws of Motion

Laws of Motion — Practice Questions