Work, Energy & Power
Q. A 1000 W heater operates for 1 hour. How much energy does it consume?
A.
3600 J
B.
1000 J
C.
3600000 J
D.
100000 J
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Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.
Correct Answer: C — 3600000 J
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Q. A 1000 W heater operates for 3 hours. How much energy does it consume?
A.
3000 J
B.
1080000 J
C.
3600000 J
D.
1000 J
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Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 3 hours = 10800 seconds. Thus, E = 1000 W * 10800 s = 10800000 J.
Correct Answer: C — 3600000 J
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Q. A 2 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
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Solution
Using energy conservation: mgh = 0.5mv², 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 2 kg ball is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv^2 = mgh. Solving gives h = v^2/(2g) = (15^2)/(2*9.8) = 11.47 m.
Correct Answer: B — 10 m
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Q. A 2 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.
40.4 m
D.
50.4 m
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Solution
Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.
Correct Answer: B — 30.4 m
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Q. A 2 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
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Solution
Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.
Correct Answer: B — 20 m
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Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?
A.
20 J
B.
40 J
C.
60 J
D.
80 J
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Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer: B — 40 J
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Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
30 m/s
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Solution
Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.
Correct Answer: B — 14 m/s
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Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
17.15 m/s
B.
12.25 m/s
C.
14.14 m/s
D.
10.0 m/s
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Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 15) = 17.15 m/s.
Correct Answer: A — 17.15 m/s
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Q. A 2 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²)
A.
39.2 J
B.
196 J
C.
78.4 J
D.
98.0 J
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Solution
Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.
Correct Answer: B — 196 J
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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 = 9.8 m/s²)
A.
14 m/s
B.
19.8 m/s
C.
20 m/s
D.
28 m/s
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Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². 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 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (Assume g = 10 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
40 m/s
Show solution
Solution
Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 10 m/s² × 20 m) = 20 m/s.
Correct Answer: C — 20 m/s
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Q. A 2 kg object is dropped from a height of 5 m. What is its potential energy at the top?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)
A.
19.6 J
B.
39.2 J
C.
49 J
D.
98 J
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Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 39.2 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
Show solution
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
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Q. A 2 kg object is dropped from a height of 5 m. What is the work done by gravity on the object just before it hits the ground?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Work done by gravity = m × g × h = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
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Q. A 2 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.
6 J
B.
9 J
C.
12 J
D.
18 J
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Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.
Correct Answer: B — 9 J
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Q. A 2 kg object is moving with a speed of 3 m/s. What is its total mechanical energy?
A.
9 J
B.
12 J
C.
15 J
D.
18 J
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Solution
Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.
Correct Answer: B — 12 J
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Q. A 2 kg object is moving with a velocity of 3 m/s. What is its momentum?
A.
3 kg·m/s
B.
6 kg·m/s
C.
9 kg·m/s
D.
12 kg·m/s
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Solution
Momentum = mass × velocity = 2 kg × 3 m/s = 6 kg·m/s.
Correct Answer: B — 6 kg·m/s
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Q. A 2 kg object is raised to a height of 5 m. What is the potential energy gained?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Potential Energy (PE) = m * g * h = 2 kg * 9.8 m/s^2 * 5 m = 98 J
Correct Answer: B — 20 J
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Q. A 2 kg object is thrown 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: KE_initial = PE_max; 0.5 × 2 kg × (10 m/s)² = 2 kg × 9.8 m/s² × h; h = 5.1 m.
Correct Answer: A — 5 m
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Q. A 2 kg object 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
Show solution
Solution
Using energy conservation: KE at bottom = PE at top; 1/2 mv^2 = mgh; h = v^2/(2g) = (10^2)/(2*10) = 5 m
Correct Answer: B — 10 m
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Q. A 2000 W heater operates for 3 hours. How much energy does it consume in kilowatt-hours?
A.
6 kWh
B.
5 kWh
C.
4 kWh
D.
3 kWh
Show solution
Solution
Energy consumed in kilowatt-hours is calculated as E = P * t. Here, P = 2000 W = 2 kW and t = 3 hours. Thus, E = 2 kW * 3 h = 6 kWh.
Correct Answer: A — 6 kWh
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Q. A 3 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.
5.1 m
B.
10.2 m
C.
15.3 m
D.
20.0 m
Show solution
Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.
Correct Answer: A — 5.1 m
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Q. A 3 kg object is dropped from a height of 12 m. What is the potential energy at the top?
A.
30 J
B.
36 J
C.
60 J
D.
120 J
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Solution
Potential energy = mgh = 3 kg × 9.8 m/s² × 12 m = 352.8 J.
Correct Answer: D — 120 J
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Q. A 3 kg object is dropped from a height of 15 m. What is the potential energy at the top?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
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Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 15 m = 441 J.
Correct Answer: D — 75 J
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Q. A 3 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
Show solution
Solution
Potential energy = mass × g × height = 3 kg × 9.8 m/s² × 5 m = 147 J.
Correct Answer: B — 30 J
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Q. A 3 kg object is lifted to a height of 4 m. What is the work done against gravity?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Work done = mgh = 3 * 9.8 * 4 = 117.6 J.
Correct Answer: B — 24 J
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Q. A 3 kg object is lifted to a height of 5 m. What is the work done against gravity? (g = 9.8 m/s²)
A.
147 J
B.
294 J
C.
441 J
D.
588 J
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Solution
Work done = mgh = 3 kg × 9.8 m/s² × 5 m = 147 J.
Correct Answer: B — 294 J
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