Laws of Motion
Q. A 1 kg ball is thrown upwards with a force of 10 N. What is the net force acting on the ball at the peak of its motion?
A.
0 N
B.
10 N
C.
5 N
D.
20 N
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Solution
At the peak, the only force acting on the ball is its weight (10 N down), so the net force is 0 N.
Correct Answer: A — 0 N
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Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (Assume g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Using the formula h = v²/(2g), h = (20 m/s)² / (2 * 10 m/s²) = 400 / 20 = 20 m.
Correct Answer: B — 30 m
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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
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Solution
Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 20 N
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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
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Solution
Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 10 N
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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
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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
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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
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Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration? (Assume no friction)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be the acceleration of the block?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg block is at rest on a horizontal surface. If a horizontal force of 30 N is applied, what will be its acceleration?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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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
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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
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Q. A 10 kg object is at rest on a surface. If a force of 30 N is applied, what is the object's acceleration assuming the friction is negligible?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, acceleration a = F/m = 30 N / 10 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 10 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
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Solution
The force acting on the object is its weight, F = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: B — 20 N
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Q. A 10 kg object is hanging at rest from a rope. What is the tension in the rope?
A.
0 N
B.
10 N
C.
100 N
D.
50 N
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Solution
The tension in the rope must balance the weight of the object. T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: C — 100 N
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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
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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
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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
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Solution
Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.
Correct Answer: A — 20 N
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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 4 m/s. What is the centripetal force acting on the object?
A.
8 N
B.
16 N
C.
32 N
D.
40 N
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Solution
Centripetal force F = mv²/r = 10 kg * (4 m/s)² / 5 m = 32 N.
Correct Answer: B — 16 N
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Q. A 10 kg object is moving with a velocity of 2 m/s. If it comes to rest, what is the change in momentum?
A.
20 kg·m/s
B.
10 kg·m/s
C.
5 kg·m/s
D.
0 kg·m/s
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Solution
Initial momentum = mv = 10 kg * 2 m/s = 20 kg·m/s. Final momentum = 0. Change in momentum = 20 kg·m/s - 0 = 20 kg·m/s.
Correct Answer: A — 20 kg·m/s
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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
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Solution
Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.
Correct Answer: B — 30 kg·m/s
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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
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Solution
Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.
Correct Answer: C — 50 kg·m/s
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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
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Solution
Net force = applied force - friction = 30 N - 10 N = 20 N.
Correct Answer: B — 20 N
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Q. A 12 kg box is pulled with a force of 60 N at an angle of 30 degrees to the horizontal. What is the horizontal component of the force?
A.
30 N
B.
60 N
C.
51.96 N
D.
20 N
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Solution
The horizontal component is F_horizontal = F * cos(θ) = 60 N * cos(30°) = 60 N * (√3/2) = 51.96 N.
Correct Answer: C — 51.96 N
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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
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Solution
The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: C — 120 N
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Q. A 12 kg object is in free fall. What is the force acting on it due to gravity?
A.
12 N
B.
24 N
C.
36 N
D.
48 N
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Solution
The force due to gravity is F = mg = 12 kg * 9.8 m/s² = 117.6 N (approximately 120 N).
Correct Answer: B — 24 N
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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its kinetic energy?
A.
150 J
B.
120 J
C.
100 J
D.
75 J
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Solution
Kinetic energy KE = 0.5 * m * v² = 0.5 * 12 kg * (5 m/s)² = 150 J.
Correct Answer: B — 120 J
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Q. A 12 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
30 kg·m/s
B.
60 kg·m/s
C.
90 kg·m/s
D.
120 kg·m/s
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Solution
Momentum p = mv = 12 kg * 5 m/s = 60 kg·m/s.
Correct Answer: B — 60 kg·m/s
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Q. A 12 kg object is pulled with a force of 24 N. If the object experiences a frictional force of 4 N, what is its acceleration?
A.
1.67 m/s²
B.
2 m/s²
C.
2.5 m/s²
D.
3 m/s²
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Solution
Net force = 24 N - 4 N = 20 N. Acceleration a = F/m = 20 N / 12 kg = 1.67 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 30 N. If the frictional force opposing the motion is 10 N, what is the acceleration of the box?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Net force = applied force - friction = 30 N - 10 N = 20 N. Acceleration a = F/m = 20 N / 15 kg = 1.33 m/s².
Correct Answer: B — 2 m/s²
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Q. A 15 kg box is pushed with a force of 45 N. What is the acceleration of the box? (Assume no friction)
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
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Solution
Using F = ma, we have a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 15 kg box is pushed with a force of 60 N on a rough surface with a frictional force of 15 N. What is the box's acceleration?
A.
3 m/s²
B.
4 m/s²
C.
5 m/s²
D.
6 m/s²
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Solution
Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 4 m/s²
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Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 20 N, what is the acceleration of the box?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Net force = applied force - friction = 60 N - 20 N = 40 N. Acceleration a = F/m = 40 N / 15 kg = 2.67 m/s².
Correct Answer: B — 3 m/s²
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