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Physics with Algebra: Work, Energy, and the Work-Energy Theorem Study Guide

스터디 가이드 - 스마트 노트

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Q1. A student pushes a box with a constant horizontal force of 50 N. The box moves 6.0 m in the direction of the force. Calculate the work done by the student on the box. (Use , )

Background

Topic: Work in Physics

This question tests your understanding of how to calculate work when a force acts in the direction of motion.

Key formula:

  • = work (in joules, J)

  • = force (in newtons, N)

  • = displacement (in meters, m)

  • = angle between force and displacement (degrees)

Step-by-Step Guidance

  1. Identify the values: , , .

  2. Recall that , so the force is fully in the direction of motion.

  3. Set up the formula: .

  4. Plug in the values for , , and .

Try solving on your own before revealing the answer!

Final Answer: 300 J

Since the force is in the direction of motion, all of it contributes to the work done.

Q2. A rope pulls a sled with a force of 80 N at an angle of 30 degrees above the horizontal. The sled moves 5.0 m horizontally. Calculate the work done by the rope.

Background

Topic: Work with Angled Forces

This question tests your ability to find the component of force in the direction of motion and calculate work.

Key formula:

  • = angle between force and direction of motion (here, )

Step-by-Step Guidance

  1. Identify the values: , , .

  2. Calculate (use calculator or known value).

  3. Set up the formula: .

  4. Plug in the values for , , and .

Try solving on your own before revealing the answer!

Final Answer: 346 J

Only the horizontal component of the force does work.

Q3. A 2.0 kg object speeds up from 3.0 m/s to 7.0 m/s. Use the work-energy theorem to find the net work done on the object.

Background

Topic: Work-Energy Theorem

This question tests your ability to relate work to changes in kinetic energy.

Key formula:

  • = mass (kg)

  • = velocity (m/s)

Step-by-Step Guidance

  1. Calculate initial kinetic energy: .

  2. Calculate final kinetic energy: .

  3. Subtract from to find .

Try solving on your own before revealing the answer!

Final Answer: 44 J

, ,

The net work equals the increase in kinetic energy.

Q4. A 1200 kg car slows from 20 m/s to rest during braking. Find the net work done on the car.

Background

Topic: Work-Energy Theorem (Negative Work)

This question tests your understanding of how work relates to a decrease in kinetic energy.

Key formula:

Step-by-Step Guidance

  1. Calculate initial kinetic energy: .

  2. Final kinetic energy is zero since the car comes to rest.

  3. Subtract from to find .

Try solving on your own before revealing the answer!

Final Answer: -240,000 J

, ,

The negative sign shows that energy was removed from the car (work done against motion).

Q5. A 15 kg backpack is lifted straight up 2.0 m at constant speed. Calculate the work done by the lifting force.

Background

Topic: Work Against Gravity

This question tests your ability to calculate work done when lifting an object vertically.

Key formula:

(since force and displacement are in the same direction)

(force needed to lift at constant speed)

  • = mass (kg)

  • = acceleration due to gravity ()

  • = vertical displacement (m)

Step-by-Step Guidance

  1. Calculate the force needed: .

  2. Set up the work formula: .

  3. Plug in the values for and .

Try solving on your own before revealing the answer!

Final Answer: 294 J

,

The lifting force does positive work equal to the change in gravitational potential energy.

Q6. A 5.0 kg box slides 4.0 m across a rough horizontal floor. The coefficient of kinetic friction is 0.20. Calculate the work done by friction.

Background

Topic: Work Done by Friction

This question tests your ability to calculate the work done by a force opposing motion.

Key formula:

(normal force for horizontal surface)

  • = coefficient of kinetic friction

  • = mass (kg)

  • = gravity ()

  • = displacement (m)

Step-by-Step Guidance

  1. Calculate normal force: .

  2. Calculate friction force: .

  3. Set up the work formula: .

Try solving on your own before revealing the answer!

Final Answer: -39.2 J

, ,

Friction does negative work, removing energy from the box.

Q7. A 60 kg rider starts from rest at the top of a frictionless roller coaster hill and drops 12 m vertically. Find the rider's speed at the bottom.

Background

Topic: Conservation of Energy

This question tests your ability to use energy conservation to relate potential energy loss to kinetic energy gain.

Key formula:

(since frictionless)

  • = mass (kg)

  • = gravity ()

  • = height (m)

  • = speed at bottom (m/s)

Step-by-Step Guidance

  1. Calculate initial potential energy: .

  2. Set .

  3. Set up .

  4. Solve for by equating and .

Try solving on your own before revealing the answer!

Final Answer: 15.3 m/s

, ,

The speed comes from converting all potential energy to kinetic energy.

Q8. A spring with spring constant 200 N/m is compressed by 0.25 m. Calculate the elastic potential energy stored in the spring.

Background

Topic: Elastic Potential Energy

This question tests your ability to use the spring energy formula.

Key formula:

  • = spring constant (N/m)

  • = compression/stretch distance (m)

Step-by-Step Guidance

  1. Identify , .

  2. Set up the formula: .

  3. Calculate and multiply by .

Try solving on your own before revealing the answer!

Final Answer: 6.25 J

The energy stored depends on both the spring constant and the square of the compression.

Q9. A force-displacement graph shows force increasing linearly from 0 N at 0 m to 40 N at 4 m, then staying at 40 N from 4 m to 8 m. Calculate the total work done.

Background

Topic: Work from Force-Displacement Graphs

This question tests your ability to find work as the area under a force vs. displacement graph.

Key formula:

Work = Area under the curve

For a triangle:

For a rectangle:

  • = base (displacement, m)

  • = height (force, N)

Step-by-Step Guidance

  1. Calculate area of triangle from 0 to 4 m: base = 4 m, height = 40 N.

  2. Calculate area of rectangle from 4 to 8 m: base = 4 m, height = 40 N.

  3. Add both areas to get total work.

Try solving on your own before revealing the answer!

Final Answer: 240 J

Triangle area: ; Rectangle area: ; Total:

The total work is the sum of the areas under the graph.

Q10. A 0.15 kg baseball moves at 40 m/s, and a 60 kg runner moves at 2.0 m/s. Compare their kinetic energies.

Background

Topic: Kinetic Energy Comparison

This question tests your ability to calculate and compare kinetic energies for different masses and speeds.

Key formula:

  • = mass (kg)

  • = velocity (m/s)

Step-by-Step Guidance

  1. Calculate for the baseball: .

  2. Calculate for the runner: .

  3. Compare the two values to see which is greater.

Try solving on your own before revealing the answer!

Final Answer: Baseball KE = 120 J; Runner KE = 120 J

Both have the same kinetic energy, even though their masses and speeds are very different.

Q11. A person pushes a lawn mower with a force of 120 N at an angle of 40 degrees below the horizontal. The mower moves 10 m horizontally. Calculate the work done by the applied force.

Background

Topic: Work with Angled Forces

This question tests your ability to find the horizontal component of a force and calculate work.

Key formula:

  • = angle below horizontal (here, )

Step-by-Step Guidance

  1. Identify , , .

  2. Calculate .

  3. Set up the formula: .

  4. Plug in the values for , , and .

Try solving on your own before revealing the answer!

Final Answer: 919 J

Only the horizontal component of the force does work.

Q12. A 20 kg sled starts from rest. It is pulled horizontally with a 100 N force for 15 m while friction exerts a 25 N force opposite the motion. Find the sled's final speed.

Background

Topic: Net Work and Kinetic Energy

This question tests your ability to find net work and use it to determine final speed.

Key formula:

(since initial speed is zero)

  • = mass (kg)

  • = final speed (m/s)

Step-by-Step Guidance

  1. Calculate work done by pulling force: .

  2. Calculate work done by friction: .

  3. Add to get net work: .

  4. Set and solve for .

Try solving on your own before revealing the answer!

Final Answer: 6.1 m/s

, , ,

The net work increases the sled's kinetic energy.

Q13. A 0.50 kg ball is thrown straight upward at 18 m/s. Ignore air resistance. Find the work done by gravity from the release point to the highest point.

Background

Topic: Work Done by Gravity

This question tests your ability to relate work done by gravity to changes in kinetic and potential energy.

Key formula:

(since gravity acts opposite to motion)

Alternatively, (where is the maximum height)

  • = mass (kg)

  • = initial speed (m/s)

Step-by-Step Guidance

  1. Calculate initial kinetic energy: .

  2. At the highest point, .

  3. Work done by gravity is .

Try solving on your own before revealing the answer!

Final Answer: -81 J

,

Gravity does negative work, removing energy as the ball rises.

Q14. A 500 kg elevator moves upward 3.0 m while speeding up from rest to 2.0 m/s. Calculate the work done by the cable on the elevator.

Background

Topic: Net Work, Gravity, and Kinetic Energy

This question tests your ability to combine work against gravity and work to increase kinetic energy.

Key formula:

  • = mass (kg)

  • = displacement (m)

  • = initial speed (m/s)

  • = final speed (m/s)

Step-by-Step Guidance

  1. Calculate .

  2. Calculate .

  3. Set .

Try solving on your own before revealing the answer!

Final Answer: 15,900 J

, ,

The cable must do enough work to lift the elevator and increase its speed.

Q15. Explain how the sign of work affects an object's kinetic energy. Include one example of positive work and one example of negative work.

Background

Topic: Work and Energy Change

This question tests your conceptual understanding of how work relates to changes in kinetic energy.

Key terms:

  • Positive work: increases kinetic energy

  • Negative work: decreases kinetic energy

Step-by-Step Guidance

  1. Recall that .

  2. Positive work means the object's speed increases (example: pushing a sled forward).

  3. Negative work means the object's speed decreases (example: friction slowing a moving box).

  4. Explain how the sign of work connects to energy change.

Try answering on your own before revealing the answer!

Final Answer:

Positive net work increases an object's kinetic energy, making it move faster. For example, when you push a sled, you do positive work and it speeds up. Negative net work decreases kinetic energy, slowing the object down. For example, friction does negative work on a sliding box, causing it to slow down.

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