Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (2020)
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A.
5 m/s
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B.
10 m/s
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C.
15 m/s
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D.
20 m/s
Solution
Using conservation of energy, PE at top = KE at bottom. mgh = 1/2 mv^2. 10 kg * 9.8 m/s^2 * 5 m = 1/2 * 10 kg * v^2. Solving gives v = 10 m/s.
Correct Answer: B — 10 m/s
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Q. A 10 kg block slides down a frictionless incline of height 5 m. What is its speed at the bottom? (g = 10 m/s²) (2020)
-
A.
10 m/s
-
B.
5 m/s
-
C.
15 m/s
-
D.
20 m/s
Solution
Using conservation of energy, PE at top = KE at bottom: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 5 m) = 10 m/s
Correct Answer: A — 10 m/s
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Q. A 10 kg box is at rest on a frictionless surface. A force of 30 N is applied. What is the final velocity after 3 seconds? (2023)
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A.
5 m/s
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B.
10 m/s
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C.
15 m/s
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D.
20 m/s
Solution
Acceleration = F/m = 30 N / 10 kg = 3 m/s². Final velocity = initial velocity + (acceleration × time) = 0 + (3 m/s² × 3 s) = 9 m/s.
Correct Answer: B — 10 m/s
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Q. A 10 kg object is at rest. A force of 30 N is applied to it. What will be its velocity after 3 seconds? (2023)
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A.
3 m/s
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B.
6 m/s
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C.
9 m/s
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D.
12 m/s
Solution
First, find acceleration: a = F/m = 30 N / 10 kg = 3 m/s². Then, using v = u + at, v = 0 + 3 m/s² × 3 s = 9 m/s.
Correct Answer: B — 6 m/s
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Q. A 10 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 10 m/s²) (2022)
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A.
200 J
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B.
100 J
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C.
50 J
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D.
300 J
Solution
Potential Energy (PE) = m * g * h = 10 kg * 10 m/s² * 20 m = 2000 J
Correct Answer: A — 200 J
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Q. A 10 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2022)
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A.
20 m/s
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B.
10 m/s
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C.
15 m/s
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D.
5 m/s
Solution
Using the formula v = √(2gh) = √(2 * 10 m/s² * 20 m) = √400 = 20 m/s.
Correct Answer: A — 20 m/s
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Q. A 10 kg object is dropped from a height of 5 m. What is its potential energy at the top? (2021)
-
A.
50 J
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B.
100 J
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C.
200 J
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D.
25 J
Solution
Potential Energy (PE) = m * g * h = 10 kg * 9.8 m/s^2 * 5 m = 490 J.
Correct Answer: B — 100 J
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Q. A 10 kg object is moving with a velocity of 4 m/s. What is its total mechanical energy if it is at a height of 5 m? (2020)
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A.
120 J
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B.
140 J
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C.
100 J
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D.
80 J
Solution
Total Mechanical Energy = Kinetic Energy + Potential Energy = (1/2 * 10 kg * (4 m/s)²) + (10 kg * 9.8 m/s² * 5 m) = 80 J + 490 J = 570 J
Correct Answer: B — 140 J
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Q. A 10 kg object is subjected to a net force of 30 N. What is the object's acceleration? (2019)
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
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D.
4 m/s²
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-ohm resistor is connected to a 20-volt battery. What is the power consumed by the resistor? (2021)
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A.
20 W
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B.
40 W
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C.
10 W
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D.
2 W
Solution
Power (P) can be calculated using the formula P = V^2 / R. Here, P = 20^2 / 10 = 400 / 10 = 40 W.
Correct Answer: B — 40 W
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Q. A 12 kg object is at rest. A force of 36 N is applied to it. What is the acceleration of the object? (2021)
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A.
1 m/s²
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B.
2 m/s²
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C.
3 m/s²
-
D.
4 m/s²
Solution
Using F = ma, a = F/m = 36 N / 12 kg = 3 m/s².
Correct Answer: C — 3 m/s²
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Q. A 12 kg object is at rest. A force of 36 N is applied to it. What will be its velocity after 3 seconds? (2020)
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A.
1 m/s
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B.
2 m/s
-
C.
3 m/s
-
D.
4 m/s
Solution
Acceleration = Force / Mass = 36 N / 12 kg = 3 m/s². Velocity = Initial velocity + (Acceleration × Time) = 0 + (3 m/s² × 3 s) = 9 m/s.
Correct Answer: C — 3 m/s
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Q. A 15 kg box is pushed with a force of 60 N. If the frictional force is 15 N, what is the acceleration of the box? (2023)
-
A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
-
D.
5 m/s²
Solution
Net force = 60 N - 15 N = 45 N. Acceleration a = F/m = 45 N / 15 kg = 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A 2 kg ball is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
-
A.
10 m/s
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B.
20 m/s
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C.
15 m/s
-
D.
5 m/s
Solution
Using energy conservation, Potential Energy = Kinetic Energy at the ground: mgh = 0.5 mv^2; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 20 m) = 20 m/s
Correct Answer: B — 20 m/s
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Q. A 2 kg mass is dropped from a height. What is the force acting on it just before it hits the ground? (2022)
-
A.
2 N
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B.
10 N
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C.
20 N
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D.
0 N
Solution
The force acting on it is its weight, F = mg = 2 kg × 10 m/s² = 20 N.
Correct Answer: B — 10 N
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Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
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A.
20 m/s
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B.
10 m/s
-
C.
15 m/s
-
D.
5 m/s
Solution
Using the formula v = √(2gh) = √(2 * 10 m/s² * 20 m) = √400 = 20 m/s
Correct Answer: A — 20 m/s
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Q. A 2 kg object is dropped from a height. What is the force acting on it just before it hits the ground? (2022)
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A.
2 N
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B.
9.8 N
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C.
19.6 N
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D.
29.4 N
Solution
The force acting on the object is its weight, F = mg = 2 kg × 9.8 m/s² = 19.6 N.
Correct Answer: C — 19.6 N
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Q. A 2 kg object is lifted to a height of 3 m. What is the work done against gravity? (g = 9.8 m/s²) (2023)
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A.
58.8 J
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B.
19.6 J
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C.
29.4 J
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D.
39.2 J
Solution
Work Done = mgh = 2 kg * 9.8 m/s² * 3 m = 58.8 J
Correct Answer: A — 58.8 J
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Q. A 2 kg object is moving with a velocity of 10 m/s. What is its momentum? (2021)
-
A.
5 kg·m/s
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B.
10 kg·m/s
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C.
15 kg·m/s
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D.
20 kg·m/s
Solution
Momentum = Mass × Velocity = 2 kg × 10 m/s = 20 kg·m/s.
Correct Answer: D — 20 kg·m/s
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Q. A 2 kg object is moving with a velocity of 4 m/s. What is its kinetic energy? (2023)
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A.
16 J
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B.
8 J
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C.
4 J
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D.
32 J
Solution
Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 2 kg * (4 m/s)^2 = 16 J.
Correct Answer: A — 16 J
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Q. A 2 kg object is moving with a velocity of 4 m/s. What is its momentum? (2020)
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A.
8 kg m/s
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B.
4 kg m/s
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C.
2 kg m/s
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D.
16 kg m/s
Solution
Momentum (p) = m * v = 2 kg * 4 m/s = 8 kg m/s
Correct Answer: A — 8 kg m/s
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Q. A 20 kg cart is pulled with a force of 80 N. If the frictional force is 20 N, what is the acceleration of the cart? (2020)
-
A.
2 m/s²
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B.
3 m/s²
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C.
4 m/s²
-
D.
5 m/s²
Solution
Net force = 80 N - 20 N = 60 N. Acceleration a = F/m = 60 N / 20 kg = 3 m/s².
Correct Answer: C — 4 m/s²
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Q. A 20 kg object is acted upon by a net force of 80 N. What is the acceleration of the object? (2023)
-
A.
2 m/s²
-
B.
3 m/s²
-
C.
4 m/s²
-
D.
5 m/s²
Solution
Acceleration = Net force / Mass = 80 N / 20 kg = 4 m/s².
Correct Answer: C — 4 m/s²
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Q. A 25 kg object is pulled with a force of 100 N. If the frictional force is 25 N, what is the acceleration of the object? (2023)
-
A.
2 m/s²
-
B.
3 m/s²
-
C.
4 m/s²
-
D.
5 m/s²
Solution
Net force = 100 N - 25 N = 75 N. Acceleration = Net force / Mass = 75 N / 25 kg = 3 m/s².
Correct Answer: C — 4 m/s²
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Q. A 3 kg object is dropped from a height of 12 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
-
A.
15 m/s
-
B.
10 m/s
-
C.
12 m/s
-
D.
20 m/s
Solution
Using conservation of energy, PE = KE: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 12 m) = 15.49 m/s
Correct Answer: A — 15 m/s
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Q. A 3 kg object is lifted to a height of 4 m. What is the gravitational potential energy gained? (2023)
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A.
120 J
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B.
60 J
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C.
30 J
-
D.
90 J
Solution
Potential Energy (PE) = m * g * h = 3 kg * 9.8 m/s^2 * 4 m = 117.6 J (approximately 120 J)
Correct Answer: A — 120 J
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Q. A 3 kg object is moving with a velocity of 6 m/s. What is its momentum? (2021)
-
A.
18 kg m/s
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B.
12 kg m/s
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C.
9 kg m/s
-
D.
6 kg m/s
Solution
Momentum (p) = mv = 3 kg * 6 m/s = 18 kg m/s
Correct Answer: A — 18 kg m/s
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Q. A 3 kg object is subjected to a net force of 12 N. What is the acceleration of the object? (2022)
-
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
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Q. A 5 kg object is moving with a constant velocity. What is the net force acting on it? (2022)
-
A.
0 N
-
B.
5 N
-
C.
10 N
-
D.
15 N
Solution
If the object is moving with constant velocity, the net force acting on it is 0 N according to Newton's first law.
Correct Answer: A — 0 N
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Q. A 5 kg object is moving with a velocity of 4 m/s. What is its momentum? (2023)
-
A.
10 kg m/s
-
B.
15 kg m/s
-
C.
20 kg m/s
-
D.
25 kg m/s
Solution
Momentum = mass × velocity = 5 kg × 4 m/s = 20 kg m/s.
Correct Answer: C — 20 kg m/s
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