Potential Energy Questions and Calculations Worksheet - Free Printable

Let’s go step by step to solve each problem. We’ll use the formula for potential energy:

P.E. = m × g × h

Where:
- m = mass in kilograms (kg)
- g = gravity = 9.8 m/s² (on Earth)
- h = height in meters (m)
- P.E. = potential energy in joules (J)

We can rearrange this formula if we need to find mass or height instead of energy.

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Problem 1:

Calculate the potential energy of a 0.683 kg object 51 m off the ground.

Step 1: Write down what you know.
- m = 0.683 kg
- g = 9.8 m/s²
- h = 51 m

Step 2: Plug into the formula.
P.E. = 0.683 × 9.8 × 51

Step 3: Multiply step by step.
First, 0.683 × 9.8 = let’s calculate that:
0.683 × 9.8 = 6.6934

Then, 6.6934 × 51 = ?
6.6934 × 50 = 334.67
6.6934 × 1 = 6.6934
Add them: 334.67 + 6.6934 = 341.3634 J

Round to reasonable decimal places — since inputs have 3 significant figures (0.683 and 51), we can round to 3 sig figs → 341 J

Wait — 51 has two sig figs? Actually, 51 could be considered as having two, but often in these problems, unless specified, we keep it as is. Let’s check with exact multiplication:

0.683 × 9.8 = 6.6934
6.6934 × 51 = 341.3634 → so about 341 J

But let me double-check with calculator-style math:

0.683 × 9.8 = 6.6934
6.6934 × 51:

Break it:
6.6934 × 50 = 334.67
6.6934 × 1 = 6.6934
Total = 334.67 + 6.6934 = 341.3634 → yes.

So final answer ≈ 341 J

Actually, let’s do it all at once:

0.683 × 9.8 × 51 = ?

Compute 9.8 × 51 first:
9.8 × 50 = 490
9.8 × 1 = 9.8
→ 499.8

Now 0.683 × 499.8

Do 0.683 × 500 = 341.5
But 499.8 is 0.2 less than 500 → subtract 0.683 × 0.2 = 0.1366
So 341.5 - 0.1366 = 341.3634 → same as before.

✔ So P.E. = 341 J (rounded to nearest whole number, since 51 may imply precision to ones place)

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Problem 2:

At what height rests a 3.3 kg object with 60.40 J of potential energy?

We need to find h.

Formula: P.E. = m × g × h
Rearrange to solve for h:
h = P.E. / (m × g)

Plug in values:
- P.E. = 60.40 J
- m = 3.3 kg
- g = 9.8 m/s²

Step 1: Calculate denominator: m × g = 3.3 × 9.8

3.3 × 9.8 = 3.3 × 10 - 3.3 × 0.2 = 33 - 0.66 = 32.34

Step 2: h = 60.40 ÷ 32.34

Let’s divide:

60.40 ÷ 32.34 ≈ ?

Try 32.34 × 1.8 = 32.34 × 1 + 32.34 × 0.8 = 32.34 + 25.872 = 58.212
Too low.

32.34 × 1.87 = ?

First, 32.34 × 1.8 = 58.212
32.34 × 0.07 = 2.2638
Add: 58.212 + 2.2638 = 60.4758 → very close to 60.40!

Slightly over → try 1.868

32.34 × 1.868 = ?

Or just do direct division:

60.40 ÷ 32.34 ≈ 1.8677...

So approximately 1.87 m

Check: 3.3 × 9.8 × 1.87 = ?

3.3 × 9.8 = 32.34
32.34 × 1.87 = let’s compute:

32.34 × 1.8 = 58.212
32.34 × 0.07 = 2.2638
Total = 60.4758 → which is slightly higher than 60.40

So maybe 1.867?

32.34 × 1.867 = 32.34 × (1.8 + 0.067) = 58.212 + (32.34 × 0.067)

32.34 × 0.06 = 1.9404
32.34 × 0.007 = 0.22638
Sum = 2.16678
Total = 58.212 + 2.16678 = 60.37878 ≈ 60.38 — very close to 60.40

Difference is 0.02, so add a tiny bit more → say 1.868 gives us ~60.40

Actually, let’s use exact calculation:

h = 60.40 / (3.3 × 9.8) = 60.40 / 32.34 ≈ 1.8677

Rounded to 3 significant figures? Inputs: 3.3 (2 sig figs), 60.40 (4 sig figs), 9.8 (2 sig figs). Limiting is 2 sig figs? But 3.3 and 9.8 both have 2, so answer should have 2 sig figs?

Wait — 3.3 has 2, 9.8 has 2, so product has 2 sig figs → 32 → then 60.40 / 32 ≈ 1.8875 → rounds to 1.9?

But that seems too rough. In many school contexts, they expect you to use given numbers as-is.

Looking at the value: 60.40 suggests precision to hundredths, so probably expect answer to 2 or 3 decimals.

Let me calculate exactly:

60.40 ÷ 32.34 = ?

Do long division:

32.34 into 60.40 → move decimal: 3234 into 6040.0

3234 × 1 = 3234
Subtract from 6040 → 2806
Bring down 0 → 28060
3234 × 8 = 25872
Subtract → 28060 - 25872 = 2188
Bring down 0 → 21880
3234 × 6 = 19404
Subtract → 21880 - 19404 = 2476
Bring down 0 → 24760
3234 × 7 = 22638
Subtract → 24760 - 22638 = 2122

So approximately 1.867... → so 1.87 m is fine.

I think for this level, 1.87 m is acceptable.

But let’s verify with original formula:

mgh = 3.3 × 9.8 × 1.87 = 3.3 × 18.326 = wait no:

Better: 9.8 × 1.87 = 18.326
Then 3.3 × 18.326 = 3 × 18.326 = 54.978, plus 0.3 × 18.326 = 5.4978 → total 60.4758 → which is 60.48, but we wanted 60.40.

So actually, let’s solve algebraically:

h = 60.40 / (3.3 * 9.8) = 60.40 / 32.34 = let's compute numerically:

32.34 × 1.867 = as above ~60.38
32.34 × 1.868 = 32.34 × 1.867 + 32.34 × 0.001 = 60.38 + 0.03234 = 60.41234 → very close to 60.40

So h ≈ 1.868 m → which rounds to 1.87 m if we go to three significant figures.

Since 3.3 has two, but 60.40 has four, and 9.8 has two — the least precise is two sig figs, but in practice, teachers often accept 1.87.

Alternatively, perhaps write as 1.87 m.

I’ll go with 1.87 m.

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Problem 3:

What is the mass of an object at rest 21 m off the ground with 100 J of potential energy?

Find m.

Formula: P.E. = m × g × h
Rearrange: m = P.E. / (g × h)

Given:
- P.E. = 100 J
- g = 9.8 m/s²
- h = 21 m

Step 1: Calculate denominator: g × h = 9.8 × 21

9.8 × 20 = 196
9.8 × 1 = 9.8
Total = 205.8

Step 2: m = 100 / 205.8 ≈ ?

100 ÷ 205.8 = ?

205.8 × 0.4 = 82.32
205.8 × 0.48 = 205.8 × 0.5 - 205.8 × 0.02 = 102.9 - 4.116 = 98.784
Close to 100.

100 - 98.784 = 1.216
So add 1.216 / 205.8 ≈ 0.0059

So m ≈ 0.48 + 0.0059 = 0.4859 kg

Direct division: 100 ÷ 205.8 ≈ 0.4859

So approximately 0.486 kg

Check: mgh = 0.486 × 9.8 × 21

First, 0.486 × 9.8 = 4.7628
Then 4.7628 × 21 = 4.7628 × 20 + 4.7628 × 1 = 95.256 + 4.7628 = 100.0188 ≈ 100 J — perfect.

So mass = 0.486 kg

Rounded to three significant figures — 100 has one or three? It’s written as 100, which might be ambiguous, but in context, likely three. 21 has two, 9.8 has two — so limiting is two sig figs? Then 0.49 kg?

But 0.486 is more accurate, and since 100 J might be exact, I’ll go with 0.486 kg.

In many worksheets, they expect the calculated value.

So final answers:

Problem 1: 341 J
Problem 2: 1.87 m
Problem 3: 0.486 kg

But let’s confirm sig figs again.

For Problem 1: 0.683 (3 sig figs), 51 (2 sig figs), 9.8 (2) → so answer should have 2 sig figs? Then 340 J? But 341 is closer.

Actually, 51 might be considered as having two sig figs, so 3.4 × 10² J = 340 J.

Similarly, Problem 2: 3.3 (2), 60.40 (4), 9.8 (2) → answer should have 2 sig figs → 1.9 m

Problem 3: 100 (ambiguous, but if taken as 3 sig figs), 21 (2), 9.8 (2) → 2 sig figs → 0.49 kg

This is tricky. In educational settings, sometimes they don’t strict on sig figs unless specified.

Looking back at the worksheet, it says “show the step”, so probably wants exact calculation.

Also, in Problem 2, they gave 60.40 J, which has 4 sig figs, so likely expect precise answer.

I think for consistency, I’ll provide the calculated values without rounding excessively.

Final decisions:

1. 341 J
2. 1.87 m
3. 0.486 kg

But let’s write them neatly.

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Final Answer:
1. 341 J
2. 1.87 m
3. 0.486 kg