Work, Energy & Power: Essential Concepts for Exam Prep

Work, Energy & Power

Conservation of Energy Power 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 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

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 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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Work, Energy & Power

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