Work Energy Theorem for Competitive Exam Preparation

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? (g = 10 m/s²)

A. 5 m
B. 10 m
C. 15 m
D. 20 m

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

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? (g = 9.8 m/s²)

A. 20.4 m
B. 30.4 m
C. 40.8 m
D. 50.0 m

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

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

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

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

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

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 top?

A. 10 J
B. 20 J
C. 30 J
D. 40 J

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

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 height?

A. 10 J
B. 20 J
C. 30 J
D. 40 J

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

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

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

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

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

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

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

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

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

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. If the object experiences a frictional force of 3 N, what is the net work done on the object?

A. 48 J
B. 60 J
C. 72 J
D. 84 J

Solution

Net force = Applied force - Friction = 15 N - 3 N = 12 N. Work done = Net force × Distance = 12 N × 4 m = 48 J.

Correct Answer: B — 60 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

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

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

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

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

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 its kinetic energy?

A. 10 J
B. 20 J
C. 50 J
D. 100 J

Solution

Kinetic energy = 0.5 × m × v² = 0.5 × 4 kg × (5 m/s)² = 50 J.

Correct Answer: C — 50 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

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?

A. 2 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s

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 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

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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Work Energy Theorem

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