This answer key provides solutions to a typical Work, Energy, and Power worksheet. Remember that specific values and problem setups might vary depending on the worksheet you're using. Always refer to your original worksheet for the exact problem statements and given values. This key offers a general approach and example solutions.
Key Concepts & Formulas
Before diving into the answers, let's refresh the core concepts and formulas:
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Work (W): Work is done when a force causes displacement of an object. The formula is:
W = Fd cos θwhere:- F = force (in Newtons)
- d = displacement (in meters)
- θ = angle between the force and displacement vectors
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Energy (E): Energy is the capacity to do work. Common forms include kinetic energy (KE) and potential energy (PE).
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Kinetic Energy (KE): The energy of motion. Formula:
KE = 1/2mv²where:- m = mass (in kilograms)
- v = velocity (in meters per second)
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Potential Energy (PE): Stored energy due to position or configuration. Gravitational potential energy is a common type:
PE = mghwhere:- m = mass (in kilograms)
- g = acceleration due to gravity (approximately 9.8 m/s²)
- h = height (in meters)
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Power (P): The rate at which work is done. Formula:
P = W/twhere:- W = work (in Joules)
- t = time (in seconds) Alternatively,
P = Fvwhere v is velocity.
Example Problem Solutions
This section provides example solutions. Adapt these solutions to your specific worksheet problems.
Problem 1: A 10 kg box is lifted 2 meters vertically. Calculate the work done.
Answer:
- F (force) = mg = (10 kg)(9.8 m/s²) = 98 N
- d (displacement) = 2 m
- θ (angle) = 0° (force and displacement are in the same direction)
- W = Fd cos θ = (98 N)(2 m) cos 0° = 196 J
Therefore, the work done is 196 Joules.
Problem 2: A 5 kg ball rolls down a hill with a velocity of 4 m/s. What is its kinetic energy?
Answer:
- m (mass) = 5 kg
- v (velocity) = 4 m/s
- KE = 1/2mv² = 1/2(5 kg)(4 m/s)² = 40 J
The kinetic energy of the ball is 40 Joules.
Problem 3: A machine does 500 J of work in 10 seconds. What is its power?
Answer:
- W (work) = 500 J
- t (time) = 10 s
- P = W/t = 500 J / 10 s = 50 W
The power of the machine is 50 Watts.
Problem 4 (More Complex): A 20 kg crate is pulled along a horizontal surface by a force of 100 N at an angle of 30° above the horizontal. If the crate is moved 5 meters, how much work is done?
Answer:
- F = 100 N
- d = 5 m
- θ = 30°
- W = Fd cos θ = (100 N)(5 m) cos 30° ≈ 433 J
The work done is approximately 433 Joules.
Important Note:
This answer key provides example solutions. Your worksheet will likely have different numerical values and potentially more complex problems. Carefully review the formulas and apply them to the specific data provided in your assignment. If you are struggling with a particular problem, try breaking it down into smaller steps, clearly identifying the known values and the unknown you are trying to solve for. Remember to always include units in your answers.
