Conservation of Energy
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
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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
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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 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
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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
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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
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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
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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 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 5 m. How much work is done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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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 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 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
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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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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?
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 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 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 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 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
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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 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 moving with a speed of 10 m/s. What is its kinetic energy?
A.
150 J
B.
300 J
C.
450 J
D.
600 J
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Solution
Kinetic energy = 0.5mv² = 0.5 * 3 * (10)² = 150 J.
Correct Answer: B — 300 J
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Q. A 3 kg object is moving with a speed of 4 m/s. What is its kinetic energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Kinetic energy = 0.5mv² = 0.5 * 3 * (4)² = 24 J.
Correct Answer: B — 24 J
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Q. A 3 kg object is moving with a speed of 4 m/s. What is its total mechanical energy?
A.
24 J
B.
32 J
C.
48 J
D.
60 J
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Solution
Total mechanical energy = KE + PE. Assuming it is at ground level, PE = 0. KE = 0.5mv² = 0.5 * 3 * (4)² = 24 J.
Correct Answer: B — 32 J
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Q. A 3 kg object is moving with a speed of 5 m/s. What is its kinetic energy?
A.
15 J
B.
25 J
C.
35 J
D.
45 J
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Solution
Kinetic energy is given by KE = 0.5mv² = 0.5 * 3 * (5)² = 37.5 J, approximately 35 J.
Correct Answer: B — 25 J
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Q. A 5 kg object is dropped from a height of 10 m. What is the total mechanical energy just before it hits the ground?
A.
0 J
B.
50 J
C.
100 J
D.
200 J
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Solution
Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 10 = 490 J. Just before hitting the ground, total energy = 490 J.
Correct Answer: C — 100 J
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Q. A 5 kg object is dropped from a height of 20 m. What is the potential energy at the top? (g = 9.8 m/s²)
A.
980 J
B.
490 J
C.
196 J
D.
9800 J
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Solution
Potential energy is given by PE = mgh. Here, PE = 5 * 9.8 * 20 = 980 J.
Correct Answer: A — 980 J
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Q. A 5 kg object is dropped from a height of 20 m. What is the total mechanical energy just before it hits the ground?
A.
0 J
B.
100 J
C.
200 J
D.
500 J
Show solution
Solution
Total mechanical energy is conserved. At height, PE = mgh = 5 * 9.8 * 20 = 980 J. Just before hitting the ground, total energy = 980 J.
Correct Answer: C — 200 J
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Q. A 5 kg object is lifted to a height of 10 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 is W = mgh = 5 * 9.8 * 10 = 490 J.
Correct Answer: D — 200 J
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