Laws of Motion

Circular Motion Friction Newtons Laws

Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?

A. 0.5 m
B. 1 m
C. 2 m
D. 0.25 m

Solution

Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.

Correct Answer: A — 0.5 m

Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the force exerted by the spring when it is compressed by 0.1 m?

A. 2 N
B. 5 N
C. 10 N
D. 20 N

Solution

Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.

Correct Answer: C — 10 N

Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when it is compressed by 0.1 m?

A. 2 N
B. 5 N
C. 20 N
D. 10 N

Solution

Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.

Correct Answer: C — 20 N

Q. A 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?

A. 6 N
B. 10 N
C. 4 N
D. 0 N

Solution

Net force = applied force - frictional force = 10 N - 4 N = 6 N.

Correct Answer: A — 6 N

Q. A 10 kg object is at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².

Correct Answer: B — 3 m/s²

Q. A 10 kg object is hanging from a rope. What is the tension in the rope when the object is at rest?

A. 0 N
B. 10 N
C. 100 N
D. 50 N

Solution

At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.

Correct Answer: C — 100 N

Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 10 m/s. What is the centripetal force acting on the object?

A. 20 N
B. 10 N
C. 5 N
D. 50 N

Solution

Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.

Correct Answer: A — 20 N

Q. A 10 kg object is moving with a velocity of 3 m/s. What is its momentum?

A. 10 kg·m/s
B. 30 kg·m/s
C. 3 kg·m/s
D. 0.3 kg·m/s

Solution

Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.

Correct Answer: B — 30 kg·m/s

Q. A 10 kg object is moving with a velocity of 5 m/s. What is its momentum?

A. 10 kg·m/s
B. 25 kg·m/s
C. 50 kg·m/s
D. 75 kg·m/s

Solution

Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.

Correct Answer: C — 50 kg·m/s

Q. A 10 kg object is pushed with a force of 30 N. If the frictional force is 10 N, what is the net force?

A. 10 N
B. 20 N
C. 30 N
D. 40 N

Solution

Net force = applied force - friction = 30 N - 10 N = 20 N.

Correct Answer: B — 20 N

Q. A 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?

A. 0 N
B. 12 N
C. 120 N
D. 100 N

Solution

The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.

Correct Answer: C — 120 N

Q. A 15 kg cart is pushed with a force of 60 N. If the frictional force opposing the motion is 15 N, what is the acceleration of the cart?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Net force = 60 N - 15 N = 45 N. Using F = ma, a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 15 kg object is in equilibrium. What is the net force acting on it?

A. 0 N
B. 15 N
C. 30 N
D. 45 N

Solution

In equilibrium, the net force acting on the object is 0 N.

Correct Answer: A — 0 N

Q. A 15 kg object is pushed with a force of 45 N. What is the acceleration of the object?

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, acceleration a = F/m = 45 N / 15 kg = 3 m/s².

Correct Answer: B — 3 m/s²

Q. A 2 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

A. 4.9 m/s²
B. 9.8 m/s²
C. 3.9 m/s²
D. 1.96 m/s²

Solution

The acceleration a = g sin(θ) = 9.8 m/s² * sin(30°) = 9.8 m/s² * 0.5 = 4.9 m/s².

Correct Answer: A — 4.9 m/s²

Q. A 2 kg object is in free fall. What is the net force acting on it?

A. 0 N
B. 2 N
C. 20 N
D. 19.6 N

Solution

The net force is equal to the weight of the object: F = mg = 2 kg * 9.8 m/s² = 19.6 N.

Correct Answer: C — 20 N

Q. A 2 kg object is pulled with a force of 8 N. What is the acceleration of the object?

A. 2 m/s²
B. 4 m/s²
C. 6 m/s²
D. 8 m/s²

Solution

Using F = ma, we find a = F/m = 8 N / 2 kg = 4 m/s².

Correct Answer: B — 4 m/s²

Q. A 2 kg object is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the object?

A. 8 kg·m/s
B. 2 kg·m/s
C. 4 kg·m/s
D. 16 kg·m/s

Solution

Momentum p = mv = 2 kg * 4 m/s = 8 kg·m/s.

Correct Answer: A — 8 kg·m/s

Q. A 25 kg object is subjected to a force of 50 N. What is its acceleration?

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Using F = ma, we find a = F/m = 50 N / 25 kg = 2 m/s².

Correct Answer: B — 2 m/s²

Q. A 3 kg block is on a table and a horizontal force of 15 N is applied. If the frictional force is 5 N, what is the acceleration of the block?

A. 2 m/s²
B. 3 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².

Correct Answer: A — 2 m/s²

Q. A 3 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block?

A. 3.9 m/s²
B. 4.9 m/s²
C. 9.8 m/s²
D. 1.5 m/s²

Solution

The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².

Correct Answer: A — 3.9 m/s²

Q. A 3 kg block is sliding on a frictionless surface with a velocity of 4 m/s. What is the momentum of the block?

A. 12 kg·m/s
B. 8 kg·m/s
C. 6 kg·m/s
D. 4 kg·m/s

Solution

Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.

Correct Answer: A — 12 kg·m/s

Q. A 3 kg object is at rest on a horizontal surface. If a force of 12 N is applied, what is the object's acceleration? (Assume no friction)

A. 2 m/s²
B. 3 m/s²
C. 4 m/s²
D. 5 m/s²

Solution

Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: C — 4 m/s²

Q. A 3 kg object is at rest on a table. If a force of 15 N is applied horizontally, what is the object's acceleration?

A. 5 m/s²
B. 10 m/s²
C. 15 m/s²
D. 20 m/s²

Solution

Using F = ma, a = F/m = 15 N / 3 kg = 5 m/s².

Correct Answer: A — 5 m/s²

Q. A 3 kg object is at rest on a table. If a horizontal force of 12 N is applied, what is the acceleration of the object?

A. 4 m/s²
B. 0 m/s²
C. 3 m/s²
D. 12 m/s²

Solution

Using F = ma, acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Correct Answer: A — 4 m/s²

Q. A 4 kg object is at rest on a table. What is the force of static friction acting on it?

A. 0 N
B. 4 N
C. 40 N
D. 10 N

Solution

If the object is at rest and no external force is applied, the static friction force is 0 N.

Correct Answer: A — 0 N

Q. A 4 kg object is at rest on a table. What is the force of static friction if no external force is applied?

A. 0 N
B. 4 N
C. 40 N
D. None of the above

Solution

If no external force is applied, the force of static friction is 0 N.

Correct Answer: A — 0 N

Q. A 4 kg object is moving with a velocity of 2 m/s. What is the change in momentum if the object comes to a stop?

A. 0 kg·m/s
B. 4 kg·m/s
C. 8 kg·m/s
D. 2 kg·m/s

Solution

Change in momentum = final momentum - initial momentum = 0 - (4 kg * 2 m/s) = -8 kg·m/s.

Correct Answer: C — 8 kg·m/s

Q. A 4 kg object is pulled with a force of 20 N. If the frictional force is 4 N, what is the net force acting on the object?

A. 16 N
B. 20 N
C. 24 N
D. 4 N

Solution

Net force = applied force - frictional force = 20 N - 4 N = 16 N.

Correct Answer: A — 16 N

Q. A 4 kg object is pushed with a force of 20 N. If the frictional force is 4 N, what is the acceleration of the object?

A. 4 m/s²
B. 5 m/s²
C. 6 m/s²
D. 7 m/s²

Solution

Net force = applied force - friction = 20 N - 4 N = 16 N. Acceleration a = F/m = 16 N / 4 kg = 4 m/s².

Correct Answer: B — 5 m/s²

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