Work, Energy & Power: Essential Concepts for Exam Prep

Work, Energy & Power

Conservation of Energy Power Work Energy Theorem

Q. A car of mass 1000 kg is moving at a speed of 20 m/s. What is its total mechanical energy assuming no friction?

A. 200,000 J
B. 400,000 J
C. 300,000 J
D. 100,000 J

Solution

Total Mechanical Energy = Kinetic Energy + Potential Energy. KE = 1/2 * m * v^2 = 1/2 * 1000 * (20^2) = 200,000 J (assuming PE = 0)

Correct Answer: B — 400,000 J

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Q. A car of mass 1000 kg is moving with a speed of 20 m/s. What is its kinetic energy?

A. 200,000 J
B. 400,000 J
C. 800,000 J
D. 1,000,000 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 1000 kg × (20 m/s)² = 200,000 J.

Correct Answer: B — 400,000 J

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the average power output if the cyclist has a mass of 70 kg?

A. 140 W
B. 280 W
C. 560 W
D. 700 W

Solution

Kinetic Energy = 0.5 × m × v² = 0.5 × 70 kg × (10 m/s)² = 3500 J. Power = Work / Time = 3500 J / 5 s = 700 W.

Correct Answer: C — 560 W

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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the work done by the cyclist?

A. 800 J
B. 900 J
C. 1000 J
D. 1200 J

Solution

Work done = Change in Kinetic Energy = 0.5 × mass × (final velocity² - initial velocity²) = 0.5 × 75 kg × (15 m/s)² = 843.75 J.

Correct Answer: C — 1000 J

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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?

A. 500 J
B. 750 J
C. 1000 J
D. 1250 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.

Correct Answer: C — 1000 J

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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the effective weight of the cyclist?

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work / Height = 300 J / 5 m = 60 N; mass = Weight / g = 60 N / 9.8 m/s² ≈ 6.12 kg.

Correct Answer: B — 60 kg

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Q. A cyclist exerts a force of 100 N to maintain a speed of 5 m/s. What is the power output of the cyclist?

A. 200 W
B. 500 W
C. 1000 W
D. 250 W

Solution

Power can be calculated using P = F * v. Here, F = 100 N and v = 5 m/s. Thus, P = 100 N * 5 m/s = 500 W.

Correct Answer: B — 500 W

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Q. A cyclist is pedaling at a constant speed and exerts a power of 200 W. If the cyclist increases their power output to 400 W, what happens to their speed assuming no other forces act?

A. Speed remains the same
B. Speed doubles
C. Speed increases by 41%
D. Speed increases by 100%

Solution

Power is proportional to the cube of the speed in cycling. If power doubles, speed increases by a factor of (2)^(1/3) which is approximately 1.26, or about a 41% increase.

Correct Answer: C — Speed increases by 41%

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Q. A force of 15 N is applied at an angle of 30° to the horizontal while moving an object 4 m. What is the work done by the force?

A. 30 J
B. 60 J
C. 90 J
D. 120 J

Solution

Work done = F × d × cos(θ) = 15 N × 4 m × cos(30°) = 15 N × 4 m × (√3/2) = 60 J.

Correct Answer: C — 90 J

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Q. A force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m. What is the work done?

A. 30 J
B. 60 J
C. 45 J
D. 75 J

Solution

Work Done = F * d * cos(θ) = 15 N * 4 m * cos(60°) = 30 J

Correct Answer: C — 45 J

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Q. A force of 15 N is applied to move an object 3 m at an angle of 60 degrees to the horizontal. What is the work done by the force?

A. 15 J
B. 20 J
C. 30 J
D. 45 J

Solution

Work done = F × d × cos(θ) = 15 N × 3 m × cos(60°) = 15 N × 3 m × 0.5 = 22.5 J.

Correct Answer: C — 30 J

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Q. A force of 15 N is applied to move an object 4 m in the direction of the force. What is the work done?

A. 30 J
B. 60 J
C. 75 J
D. 90 J

Solution

Work done = Force × Distance = 15 N × 4 m = 60 J.

Correct Answer: D — 90 J

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Q. A force of 25 N is applied at an angle of 30 degrees to the horizontal while moving an object 10 m. What is the work done?

A. 100 J
B. 125 J
C. 150 J
D. 175 J

Solution

Work done = Force × Distance × cos(θ) = 25 N × 10 m × cos(30°) = 216.5 J.

Correct Answer: B — 125 J

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Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?

A. 37.5 J
B. 50 J
C. 75 J
D. 100 J

Solution

Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.

Correct Answer: A — 37.5 J

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Q. A light bulb uses 60 W of power. How much energy does it consume in 1 hour?

A. 3600 J
B. 216000 J
C. 60000 J
D. 180000 J

Solution

Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 1 hour = 3600 seconds. Thus, E = 60 W * 3600 s = 216000 J.

Correct Answer: B — 216000 J

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Q. A light bulb uses 60 W of power. How much energy does it consume in 2 hours?

A. 120 J
B. 7200 J
C. 432000 J
D. 360 J

Solution

Energy consumed can be calculated using E = P * t. Here, P = 60 W and t = 2 hours = 7200 seconds. Thus, E = 60 W * 7200 s = 432000 J.

Correct Answer: C — 432000 J

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Q. A light bulb uses 60 W of power. How much energy does it consume in 5 hours?

A. 18000 J
B. 108000 J
C. 300000 J
D. 360000 J

Solution

Energy consumed can be calculated using E = P * t. Here, E = 60 W * (5 * 3600 s) = 108000 J.

Correct Answer: B — 108000 J

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Q. A light bulb uses 60 Watts of power. How much energy does it consume in 2 hours?

A. 120 Joules
B. 7200 Joules
C. 432000 Joules
D. 360000 Joules

Solution

Energy consumed = Power × Time = 60 W × (2 × 3600 s) = 432000 J.

Correct Answer: C — 432000 Joules

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Q. A machine does 300 J of work in 5 seconds. What is its power?

A. 60 W
B. 30 W
C. 50 W
D. 20 W

Solution

Power (P) = Work / Time = 300 J / 5 s = 60 W

Correct Answer: A — 60 W

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Q. A machine does 500 J of work in 10 seconds. What is its power output?

A. 50 W
B. 100 W
C. 200 W
D. 500 W

Solution

Power is calculated using the formula P = W/t. Here, W = 500 J and t = 10 s, so P = 500 J / 10 s = 50 W.

Correct Answer: B — 100 W

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Q. A machine does 500 J of work in 10 seconds. What is its power?

A. 50 W
B. 100 W
C. 200 W
D. 500 W

Solution

Power = Work / Time = 500 J / 10 s = 50 W.

Correct Answer: B — 100 W

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Q. A machine does 600 J of work in 5 seconds. What is its power?

A. 120 W
B. 100 W
C. 150 W
D. 200 W

Solution

Power (P) = Work / Time = 600 J / 5 s = 120 W

Correct Answer: A — 120 W

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Q. A motor has a power rating of 1500 Watts. How much work can it do in 10 minutes?

A. 90000 Joules
B. 15000 Joules
C. 25000 Joules
D. 30000 Joules

Solution

Work done = Power × Time = 1500 W × (10 × 60 s) = 90000 J.

Correct Answer: A — 90000 Joules

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Q. A particle moves in a straight line under the influence of a constant force of 12 N. If the particle moves 3 m, what is the work done by the force?

A. 12 J
B. 24 J
C. 36 J
D. 48 J

Solution

Work done = Force × Distance = 12 N × 3 m = 36 J.

Correct Answer: B — 24 J

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Q. A particle moves in a straight line under the influence of a constant force of 3 N. If it moves 6 m, what is the work done by the force?

A. 9 J
B. 12 J
C. 15 J
D. 18 J

Solution

Work done = Force × Distance = 3 N × 6 m = 18 J.

Correct Answer: D — 18 J

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Q. A particle moves in a straight line under the influence of a constant force. If the initial kinetic energy is 100 J and the work done by the force is 50 J, what is the final kinetic energy?

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Final kinetic energy = Initial kinetic energy + Work done = 100 J + 50 J = 150 J.

Correct Answer: C — 150 J

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Q. A pendulum of length 2 m swings from a height of 1 m. What is the speed at the lowest point of the swing? (g = 9.8 m/s²)

A. 4.4 m/s
B. 3.1 m/s
C. 2.8 m/s
D. 5.0 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.

Correct Answer: A — 4.4 m/s

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Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?

A. 2 m/s
B. 4 m/s
C. 6 m/s
D. 8 m/s

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.

Correct Answer: B — 4 m/s

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Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?

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

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.

Correct Answer: B — 10 m/s

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Q. A person lifts a box of mass 10 kg to a height of 2 m in 4 seconds. What is the power exerted by the person?

A. 50 W
B. 25 W
C. 100 W
D. 75 W

Solution

The work done against gravity is W = mgh = 10 kg * 9.8 m/s² * 2 m = 196 J. Power is P = W/t = 196 J / 4 s = 49 W, approximately 50 W.

Correct Answer: A — 50 W

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Work, Energy & Power

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