Work, Energy & Power
Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.
Correct Answer: B — 10 m
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Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
Show solution
Solution
Using energy conservation, mgh = 0.5 mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.
Correct Answer: A — 5 m
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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?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
Show solution
Solution
Using energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.
Correct Answer: B — 20 m
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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? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
50 m
Show solution
Solution
Using energy conservation: Initial K.E. = Potential Energy at max height. 0.5mv² = mgh. h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.
Correct Answer: B — 30 m
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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? (g = 9.8 m/s²)
A.
20.4 m
B.
30.4 m
C.
40.8 m
D.
50.0 m
Show solution
Solution
Using energy conservation, initial kinetic energy = mgh; 0.5 × 1 kg × (20 m/s)² = 9.8 m × h; h = 20.4 m.
Correct Answer: A — 20.4 m
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Q. A 1 kg mass is attached to a spring and compressed by 0.2 m. If the spring constant is 100 N/m, what is the potential energy stored in the spring?
A.
2 J
B.
4 J
C.
6 J
D.
8 J
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Solution
Potential energy in a spring is given by PE = 0.5kx². Thus, PE = 0.5 * 100 * (0.2)² = 2 J.
Correct Answer: B — 4 J
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Q. A 1 kg mass is dropped from a height of 1 m. What is its speed just before it hits the ground?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
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Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 1) = 4.43 m/s.
Correct Answer: B — 2 m/s
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Q. A 1 kg mass is dropped from a height of 1 m. What is the speed just before it hits the ground?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
Show solution
Solution
Using conservation of energy, v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.43 m/s.
Correct Answer: B — 2 m/s
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Q. A 1 kg mass is dropped from a height of 10 m. What is its speed just before it hits the ground?
A.
5 m/s
B.
10 m/s
C.
14 m/s
D.
20 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 10) = 14 m/s.
Correct Answer: C — 14 m/s
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Q. A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.
Correct Answer: B — 10 m/s
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Q. A 1 kg mass is lifted to a height of 5 m. How much work is done against gravity?
A.
5 J
B.
10 J
C.
15 J
D.
20 J
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Solution
Work done against gravity = mgh = 1 * 9.8 * 5 = 49 J.
Correct Answer: B — 10 J
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Q. A 1 kg mass is lifted vertically 10 m. How much work is done against gravity? (g = 9.8 m/s²)
A.
9.8 J
B.
19.6 J
C.
29.4 J
D.
39.2 J
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Solution
Work done = mgh = 1 kg × 9.8 m/s² × 10 m = 98 J.
Correct Answer: B — 19.6 J
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Q. A 1 kg object is pushed with a force of 10 N over a distance of 3 m. What is the work done?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Work done = Force × Distance = 10 N × 3 m = 30 J.
Correct Answer: C — 30 J
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Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?
A.
100 J
B.
200 J
C.
300 J
D.
400 J
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Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: B — 200 J
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Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.
Correct Answer: D — 28 m/s
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Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?
A.
100 J
B.
200 J
C.
300 J
D.
400 J
Show solution
Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: D — 400 J
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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground?
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 20 m/s.
Correct Answer: C — 20 m/s
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Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.
Correct Answer: B — 14 m/s
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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 = 9.8 m/s²)
A.
14 m/s
B.
19.8 m/s
C.
20 m/s
D.
28 m/s
Show solution
Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Correct Answer: B — 19.8 m/s
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Q. A 10 kg object is lifted to a height of 10 m. How much work is done against gravity?
A.
0 J
B.
100 J
C.
200 J
D.
1000 J
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Solution
Work done against gravity is equal to the change in potential energy, W = mgh = 10 * 9.8 * 10 = 980 J.
Correct Answer: D — 1000 J
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Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?
A.
20 J
B.
40 J
C.
60 J
D.
80 J
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Solution
Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.
Correct Answer: B — 40 J
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Q. A 10 kg object is lifted to a height of 5 m. How much work is done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer: B — 100 J
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Q. A 10 kg object is lifted to a height of 5 m. What is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done against gravity = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer: B — 100 J
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Q. A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?
A.
45 J
B.
90 J
C.
135 J
D.
180 J
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Solution
Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.
Correct Answer: B — 90 J
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Q. A 10 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.
45 J
B.
30 J
C.
60 J
D.
90 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer: A — 45 J
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Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?
A.
125 J
B.
250 J
C.
500 J
D.
1000 J
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Solution
Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.
Correct Answer: B — 250 J
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
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Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer: C — 45 J
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is the total mechanical energy if it is at a height of 5 m?
A.
150 J
B.
180 J
C.
300 J
D.
450 J
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Solution
Total Mechanical Energy = Kinetic Energy + Potential Energy = 0.5 × 10 kg × (3 m/s)² + 10 kg × 9.8 m/s² × 5 m = 45 J + 490 J = 535 J.
Correct Answer: C — 300 J
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Q. A 10 kg object is moving with a velocity of 4 m/s. What is its momentum?
A.
40 kg m/s
B.
20 kg m/s
C.
10 kg m/s
D.
30 kg m/s
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Solution
Momentum (p) = m * v = 10 kg * 4 m/s = 40 kg m/s
Correct Answer: A — 40 kg m/s
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Q. A 10 kg object is thrown upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.
11.5 m
B.
22.5 m
C.
15.3 m
D.
10.0 m
Show solution
Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.
Correct Answer: A — 11.5 m
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