Physics practice worksheet focusing on work, energy, and power calculations.

Let’s solve each problem step by step, using the fundamental physics formulas for work, gravitational potential energy, and understanding when no work is done.

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Problem 1:

> How far can a mother push a 20.0 kg baby carriage, using a force of 62 N, if she can only do 2920 J of work?

Formula:
Work = Force × Distance
→ \( W = F \cdot d \)
→ \( d = \frac{W}{F} \)

Given:
- \( W = 2920 \, \text{J} \)
- \( F = 62 \, \text{N} \)

Solution:
\[
d = \frac{2920}{62} \approx 47.1 \, \text{m}
\]

✔ Answer: 47.1 meters

*(Note: The mass of the carriage (20.0 kg) is irrelevant here because we’re given the applied force directly — no need to calculate force from weight or friction unless specified.)*

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Problem 2:

> Calculate the work done by a 47 N force pushing a pencil 0.26 m.

Formula:
\( W = F \cdot d \)

Given:
- \( F = 47 \, \text{N} \)
- \( d = 0.26 \, \text{m} \)

Solution:
\[
W = 47 \times 0.26 = 12.22 \, \text{J}
\]

✔ Answer: 12.22 J

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Problem 3:

> A girl pushes her little brother on his sled with a force of 300 N for 750 m. How much work is this if the force of friction acting on the sled is (a) 200 N, (b) 300 N?

Important Concept:
Work done *by the girl* is based on the force she applies and the distance moved in the direction of that force — not net force or friction, unless asked for *net work*.

So regardless of friction, if she applies 300 N over 750 m, the work she does is:

\[
W = 300 \, \text{N} \times 750 \, \text{m} = 225,000 \, \text{J}
\]

Friction affects the *net work* or acceleration, but not the work *she does*.

✔ Answer: (a) 225,000 J, (b) 225,000 J

*(If the question had asked “net work” or “useful work”, then we’d subtract friction. But it says “how much work is this” — meaning work done by her.)*

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Problem 4:

> A 75.0 kg man pushes on a 500,000 t wall for 250 s but it does not move. How much work does he do on the wall?

Key Concept:
Work requires displacement in the direction of the force.

If the wall doesn’t move → displacement = 0 → Work = 0

Formula:
\( W = F \cdot d \), if \( d = 0 \), then \( W = 0 \)

✔ Answer: 0 J

*(Even though he exerts force and spends time/energy, no mechanical work is done on the wall. His muscles may do internal work, but that’s biological, not physics work.)*

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Problem 5:

> A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min using a force of 40 N. How much work has the boy done?

Steps:
1. Convert time to seconds:
\( 30 \, \text{min} = 30 \times 60 = 1800 \, \text{s} \)

2. Find distance:
\( d = v \cdot t = 0.80 \, \text{m/s} \times 1800 \, \text{s} = 1440 \, \text{m} \)

3. Calculate work:
\( W = F \cdot d = 40 \, \text{N} \times 1440 \, \text{m} = 57,600 \, \text{J} \)

✔ Answer: 57,600 J

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Problem 6:

> What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building relative to (a) the tenth floor, (b) the sixth floor, (c) the first floor. (Each storey is 2.50 m high.)

Formula:
Gravitational Potential Energy:
\( PE = mgh \)

Where:
- \( m = 61.2 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h \) = height above reference point

Height of roof:
10 storeys × 2.50 m = 25.0 m above ground.

(a) Relative to the 10th floor:
→ Height difference = 0 m
→ \( PE = 61.2 \times 9.8 \times 0 = 0 \, \text{J} \)

(b) Relative to the 6th floor:
→ Height difference = (10 - 6) × 2.50 = 4 × 2.50 = 10.0 m
→ \( PE = 61.2 \times 9.8 \times 10.0 = 6000 \, \text{J} \) *(rounded to 3 sig figs)*
Exact: \( 61.2 × 9.8 × 10 = 6000.0 - wait, let’s compute:*

\[
61.2 × 9.8 = 600.0 - actually 61.2 × 9.8 = 600.0? Let's calculate:
61.2 × 9.8 = 61.2 × (10 - 0.2) = 612 - 12.24 = 599.76
Then × 10 = 5997.6 ≈ 6000 J \, (\text{to 3 sig figs})
\]

(c) Relative to the 1st floor:
→ Height difference = (10 - 1) × 2.50 = 9 × 2.50 = 22.5 m
→ \( PE = 61.2 × 9.8 × 22.5 \)

First: 61.2 × 9.8 = 599.76
Then: 599.76 × 22.5 = ?

Break it down:
599.76 × 20 = 11,995.2
599.76 × 2.5 = 1,499.4
Total = 11,995.2 + 1,499.4 = 13,494.6 J ≈ 13,500 J (to 3 sig figs)

✔ Answers:
(a) 0 J
(b) 6000 J
(c) 13,500 J

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Problem 7:

> A 10,000 kg airplane lands, descending a vertical distance of 10 km while travelling 100 km measured along the ground. What is the plane’s loss of potential energy?

Key Point:
Potential energy depends only on vertical height change, not horizontal distance.

Formula:
Loss in PE = \( mgh \)

Given:
- \( m = 10,000 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 10 \, \text{km} = 10,000 \, \text{m} \)

Solution:
\[
\Delta PE = 10,000 × 9.8 × 10,000 = 980,000,000 \, \text{J} = 9.8 × 10^8 \, \text{J}
\]

✔ Answer: 9.8 × 10⁸ J (or 980 MJ)

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Problem 8:

> Calculate the work done by a 2.4 N force pushing a 400 g sandwich across a table 0.75 m wide.

Formula:
\( W = F \cdot d \)

Given:
- \( F = 2.4 \, \text{N} \)
- \( d = 0.75 \, \text{m} \)

Mass (400 g = 0.4 kg) is irrelevant — unless friction is involved, which it isn’t mentioned.

Solution:
\[
W = 2.4 × 0.75 = 1.8 \, \text{J}
\]

✔ Answer: 1.8 J

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## ✔ Final Answers Summary:

1. 47.1 m
2. 12.22 J
3. (a) 225,000 J, (b) 225,000 J
4. 0 J
5. 57,600 J
6. (a) 0 J, (b) 6000 J, (c) 13,500 J
7. 9.8 × 10⁸ J
8. 1.8 J

Let me know if you’d like diagrams or more explanation on any part!