Work Energy Theorem
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
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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 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, 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 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
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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 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 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 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 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
Show solution
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 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
Show solution
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 5 m. What is the potential energy at the height?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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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 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 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
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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 moving with a speed of 4 m/s. If a net work of 24 J is done on the object, what will be its final speed?
A.
4 m/s
B.
6 m/s
C.
8 m/s
D.
10 m/s
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Solution
Initial kinetic energy = 0.5 × m × v^2 = 0.5 × 3 kg × (4 m/s)^2 = 24 J. Final kinetic energy = Initial + Work done = 24 J + 24 J = 48 J. Final speed = √(2 × KE/m) = √(2 × 48 J / 3 kg) = 6.93 m/s.
Correct Answer: C — 8 m/s
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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 12 N × 4 m = 48 J. Kinetic energy = 0.5 × mass × v²; 48 J = 0.5 × 3 kg × v²; v² = 32; v = 4 m/s.
Correct Answer: C — 4 m/s
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Q. A 3 kg object is pushed with a force of 12 N over a distance of 4 m. What is the net work done if friction does 8 J of work?
A.
28 J
B.
32 J
C.
36 J
D.
40 J
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Solution
Net work done = Work done by force - Work done against friction = (12 N × 4 m) - 8 J = 48 J - 8 J = 40 J.
Correct Answer: B — 32 J
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Q. A 3 kg object is pushed with a force of 15 N over a distance of 4 m. What is the work done on the object?
A.
30 J
B.
45 J
C.
60 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer: C — 60 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the change in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Change in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer: B — 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Increase in potential energy = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer: B — 24 J
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Q. A 4 kg object is lifted to a height of 3 m. What is the increase in gravitational potential energy? (g = 9.8 m/s²)
A.
117.6 J
B.
117 J
C.
120 J
D.
150 J
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Solution
Potential energy increase = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer: A — 117.6 J
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Q. A 4 kg object is moving with a speed of 5 m/s. If it comes to rest, what is the work done by friction?
A.
50 J
B.
75 J
C.
100 J
D.
125 J
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Solution
Work done = change in kinetic energy = 0 - 0.5 × 4 kg × (5 m/s)² = -50 J.
Correct Answer: C — 100 J
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Q. A 4 kg object is moving with a speed of 5 m/s. What is the total mechanical energy if it is at a height of 2 m?
A.
50 J
B.
60 J
C.
70 J
D.
80 J
Show solution
Solution
Total mechanical energy = Kinetic energy + Potential energy = 0.5 × 4 kg × (5 m/s)² + 4 kg × 9.8 m/s² × 2 m = 50 J + 78.4 J = 128.4 J.
Correct Answer: C — 70 J
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed? (Assume no friction)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. Kinetic energy = 0.5 × m × v². 60 J = 0.5 × 4 kg × v². v² = 30, v = √30 ≈ 5.48 m/s.
Correct Answer: C — 4 m/s
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Q. A 4 kg object is pushed with a force of 20 N over a distance of 3 m. If the object starts from rest, what is its final speed?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Work done = Force × Distance = 20 N × 3 m = 60 J. K.E = 0.5 × m × v², 60 J = 0.5 × 4 kg × v², v = 3.87 m/s.
Correct Answer: C — 4 m/s
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Q. A block of mass 2 kg is pushed along a frictionless surface by a constant force of 10 N. What is the work done by the force when the block moves 5 m?
A.
10 J
B.
20 J
C.
30 J
D.
50 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
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Q. A block of mass 2 kg is pushed along a frictionless surface by a force of 10 N. What is the work done by the force when the block moves 5 m?
A.
10 J
B.
20 J
C.
30 J
D.
50 J
Show solution
Solution
Work done = Force × Distance = 10 N × 5 m = 50 J.
Correct Answer: B — 20 J
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