Potential and Kinetic Energy Worksheet - Practice Problems

Quick Look

This worksheet helps you practice identifying and calculating potential and kinetic energy. You will decide whether an object has PE, KE, or both. Then you will use the formulas to calculate the energy. Check your answers with the conservation of energy rule.

Section 1: Identify the Energy Type

Read each description. Write PE (potential energy), KE (kinetic energy), or both.

1. A rock sitting on the edge of a cliff. ____
2. A child sliding down a slide. ____
3. A stretched rubber band waiting to be released. ____
4. A car driving down the highway. ____
5. A book on the top shelf of a bookcase. ____
6. A basketball bouncing on the court. ____
7. An arrow held in a drawn bow. ____
8. A roller coaster car at the top of a hill. ____
9. A wind turbine spinning in the breeze. ____
10. A diver standing on a diving board. ____

Answers: 1. PE, 2. KE, 3. PE, 4. KE, 5. PE, 6. Both (has KE from moving and elastic PE from compressing), 7. PE (elastic), 8. PE, 9. KE (rotational), 10. PE

Section 2: Calculate Kinetic Energy

Use the formula KE = 1/2mv squared to find the kinetic energy. Show your work.

1. A 3 kg ball rolls at 4 m/s. What is its KE?

2. A 40 kg child runs at 5 m/s. Find the KE.

3. A 0.15 kg baseball is thrown at 30 m/s. Calculate the KE.

4. A 1,200 kg car travels at 20 m/s. How much KE does it have?

5. A 0.5 kg bird flies at 10 m/s. What is its KE?

Answers:

1. KE = 1/2 x 3 x 16 = 24 J
2. KE = 1/2 x 40 x 25 = 500 J
3. KE = 1/2 x 0.15 x 900 = 67.5 J
4. KE = 1/2 x 1,200 x 400 = 240,000 J
5. KE = 1/2 x 0.5 x 100 = 25 J

Section 3: Calculate Potential Energy

Use the formula PE = mgh to find the gravitational potential energy. Use g = 9.8 m/s squared.

1. A 5 kg box sits on a 3 m high shelf. What is its PE?

2. A 60 kg person stands on a 10 m high diving platform. Find the PE.

3. A 2 kg book is on a 2 m high table. Calculate the PE.

4. A 500 kg roller coaster car is at the top of a 50 m hill. What is its PE?

5. A 0.2 kg apple hangs 4 m above the ground. Find the PE.

Answers:

1. PE = 5 x 9.8 x 3 = 147 J
2. PE = 60 x 9.8 x 10 = 5,880 J
3. PE = 2 x 9.8 x 2 = 39.2 J
4. PE = 500 x 9.8 x 50 = 245,000 J
5. PE = 0.2 x 9.8 x 4 = 7.84 J

Section 4: Energy Conversion Problems

Use the conservation of mechanical energy. PE + KE stays the same (ignore friction).

1. A 2 kg ball is dropped from 5 m. How much KE does it have just before hitting the ground?

   Solution: PE at top = 2 x 9.8 x 5 = 98 J KE at top = 0 J Total E = 98 J Just before ground: PE = 0 J, so KE = 98 J

2. A 10 kg rock falls from 20 m. What is its speed just before impact?

   Solution: PE at top = 10 x 9.8 x 20 = 1,960 J KE at bottom = 1,960 J v squared = 2 x 1,960 / 10 = 392 v = 19.8 m/s

3. A 1 kg pendulum bob is raised 0.5 m. How fast is it going at the bottom of the swing?

   Solution: PE at top = 1 x 9.8 x 0.5 = 4.9 J KE at bottom = 4.9 J v squared = 2 x 4.9 / 1 = 9.8 v = 3.13 m/s

4. A 70 kg skier starts from rest at the top of a 30 m hill. What is their speed at the bottom?

   Solution: PE at top = 70 x 9.8 x 30 = 20,580 J KE at bottom = 20,580 J v squared = 2 x 20,580 / 70 = 588 v = 24.25 m/s

5. A 0.5 kg ball is thrown straight up at 15 m/s. How high does it go?

   Solution: KE at start = 1/2 x 0.5 x 225 = 56.25 J PE at top = 56.25 J h = 56.25 / (0.5 x 9.8) = 56.25 / 4.9 h = 11.48 m

Section 5: Harder Problems

These problems combine multiple steps.

1. A 1,500 kg car is moving at 25 m/s. The driver brakes and the car slows to 10 m/s. How much KE was removed?

   Solution: KE initial = 1/2 x 1,500 x 625 = 468,750 J KE final = 1/2 x 1,500 x 100 = 75,000 J KE removed = 468,750 - 75,000 = 393,750 J

2. A 2 kg ball is at a height of 10 m. It is dropped. At what height does it have equal PE and KE?

   Solution: Total E = 2 x 9.8 x 10 = 196 J At equal PE and KE: PE = KE = 98 J h = PE / mg = 98 / (2 x 9.8) = 98 / 19.6 = 5 m At 5 meters, PE = KE = 98 J.

3. A spring with k = 200 N/m is compressed by 0.3 m. How much elastic PE is stored?

   Solution: PE elastic = 1/2 x 200 x (0.3 x 0.3) PE elastic = 1/2 x 200 x 0.09 PE elastic = 9 J

4. A 0.1 kg ball is launched from a spring. The spring (k = 500 N/m) is compressed 0.1 m. How fast does the ball leave the spring?

   Solution: PE elastic = 1/2 x 500 x (0.1 x 0.1) PE elastic = 1/2 x 500 x 0.01 PE elastic = 2.5 J All elastic PE becomes KE: KE = 2.5 J v squared = 2 x 2.5 / 0.1 = 50 v = 7.07 m/s

Section 6: Identify the Energy at Each Point

A roller coaster car starts at point A at the top of a 40 m hill. It goes down to point B at the bottom, then up to point C at 15 m.

Fill in the energy values for a 1,000 kg car (with track and car combined).

Solution:

Point A: PE = 1,000 x 9.8 x 40 = 392,000 J KE = 0 J Total = 392,000 J

Point B: PE = 0 J KE = 392,000 J Total = 392,000 J

Point C: PE = 1,000 x 9.8 x 15 = 147,000 J KE = 392,000 - 147,000 = 245,000 J Total = 392,000 J

Teacher Corner

Discussion Questions

1. How does the conservation of energy help you check your worksheet answers?
2. Why does mass cancel out in falling object problems?
3. What happens to the total energy when friction is present?

Classroom Activity: Worksheet Race

Have students work through the worksheet in pairs. The first pair to correctly complete all sections wins. Review the answers as a class and discuss the most common mistakes.

Common Misconceptions

Some students think PE and KE are independent. They are connected. One turns into the other.

Some students think heavy objects have more KE when falling from the same height. They have more PE at the start, but the speed at the bottom is the same regardless of mass.

Some students think the total energy changes during a fall. It does not. Only the split between PE and KE changes.

References

1. U.S. Department of Energy — Office of Energy Efficiency & Renewable Energy
2. Encyclopaedia Britannica — Energy
3. Wikipedia — Energy
4. U.S. Energy Information Administration — Energy Kids
5. NASA — Earth Observatory: Energy

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Last updated: June 15, 2026

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