Physics practice worksheet featuring 11 problems focused on calculating work, power, and energy in various scenarios.

Let's solve each of these physics problems step by step. We'll use the fundamental concepts of work, power, and energy.

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

Formula for work:
$$
W = F \cdot d \cdot \cos\theta
$$
Since the force is in the direction of motion, $\theta = 0^\circ$, so $\cos\theta = 1$.

$$
W = 47\, \text{N} \times 0.26\, \text{m} = 12.22\, \text{J}
$$

✔ Answer: 12.22 J

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2. Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m against a force of 23 N.

This one is tricky. The pencil is pushed with a 47 N force, but there’s an opposing force of 23 N. However, the net force determines acceleration, but work is done by the applied force, not the net force — unless specified otherwise.

But the problem says "pushing... against a force", which implies that the applied force is 47 N, and it moves the object 0.25 m. So we calculate the work done by the 47 N force.

$$
W = F \cdot d = 47\, \text{N} \times 0.25\, \text{m} = 11.75\, \text{J}
$$

Note: The 23 N opposing force does negative work, but the question asks for the work done *by* the 47 N force.

✔ Answer: 11.75 J

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

Mass of sandwich: $0.4\, \text{kg}$ (but not needed since force is given)

Force = 2.4 N, distance = 0.75 m

$$
W = F \cdot d = 2.4\, \text{N} \times 0.75\, \text{m} = 1.8\, \text{J}
$$

✔ Answer: 1.8 J

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4. 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?

Use:
$$
W = F \cdot d \Rightarrow d = \frac{W}{F}
$$

$$
d = \frac{2920\, \text{J}}{62\, \text{N}} = 47.1\, \text{m}
$$

✔ Answer: 47.1 meters

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5. How much work is it to lift a 20 kg sack of potatoes vertically 6.5 m?

Work against gravity:
$$
W = mgh
$$
where $g = 9.8\, \text{m/s}^2$

$$
W = 20\, \text{kg} \times 9.8\, \text{m/s}^2 \times 6.5\, \text{m} = 1274\, \text{J}
$$

✔ Answer: 1274 J

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6. If a small motor does 520 J of work to move a toy 260 m, what force does it exert?

$$
W = F \cdot d \Rightarrow F = \frac{W}{d} = \frac{520\, \text{J}}{260\, \text{m}} = 2\, \text{N}
$$

✔ Answer: 2 N

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7. 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 is (a) 200 N, (b) 300 N?

Wait — this wording is ambiguous. Let's interpret carefully.

She applies a force of 300 N over 750 m. But friction opposes motion.

However, work done by the girl depends on her applied force, not the net force, unless otherwise stated.

So:

(a) Friction = 200 N
The girl applies 300 N → work done by her:
$$
W = 300\, \text{N} \times 750\, \text{m} = 225,000\, \text{J}
$$

(b) Friction = 300 N
Same logic: she still applies 300 N → same work:
$$
W = 300\, \text{N} \times 750\, \text{m} = 225,000\, \text{J}
$$

⚠️ But if the sled doesn't move (if friction equals applied force), then no displacement → work = 0. But here, the problem says she pushes for 750 m, so it does move.

So even if friction is 300 N, as long as she applies 300 N and it moves, work is still:
$$
W = F \cdot d = 300 \times 750 = 225,000\, \text{J}
$$

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

> Note: Work done by the person is based on her force and displacement, regardless of friction (unless asking for net work).

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

If the wall does not move, then displacement = 0.

$$
W = F \cdot d = F \cdot 0 = 0\, \text{J}
$$

Even though he exerts force, no work is done because there is no displacement.

✔ Answer: 0 J

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9. 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 is the boy doing?

First, find distance traveled:

Time = 30 min = $30 \times 60 = 1800\, \text{s}$

Speed = 0.80 m/s

$$
d = v \cdot t = 0.80\, \text{m/s} \times 1800\, \text{s} = 1440\, \text{m}
$$

Now work:

$$
W = F \cdot d = 40\, \text{N} \times 1440\, \text{m} = 57,600\, \text{J}
$$

✔ Answer: 57,600 J

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10. 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)

Height of each floor = 2.50 m

Total height from ground to roof (10th floor):
$$
h = 10 \times 2.50 = 25.0\, \text{m}
$$

(a) Relative to tenth floor:
He is on the 10th floor → height difference = 0 →
$$
PE = mgh = 61.2 \times 9.8 \times 0 = 0\, \text{J}
$$

(b) Relative to sixth floor:
Height difference = $ (10 - 6) \times 2.50 = 4 \times 2.50 = 10.0\, \text{m} $

$$
PE = 61.2 \times 9.8 \times 10.0 = 5,997.6\, \text{J} \approx 5998\, \text{J}
$$

(c) Relative to first floor:
Height = 25.0 m

$$
PE = 61.2 \times 9.8 \times 25.0 = 14,994\, \text{J} \approx 14,990\, \text{J}
$$

✔ Answers:
- (a) 0 J
- (b) 5,998 J
- (c) 14,990 J

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

Only vertical distance matters for gravitational potential energy.

$$
\Delta PE = mgh
$$
Here, $h = 10\, \text{m}$ (downward → loss)

$$
\Delta PE = 10,000\, \text{kg} \times 9.8\, \text{m/s}^2 \times 10\, \text{m} = 980,000\, \text{J}
$$

So, loss of potential energy = 980,000 J

✔ Answer: 980,000 J

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 12.22 J |
| 2 | 11.75 J |
| 3 | 1.8 J |
| 4 | 47.1 m |
| 5 | 1274 J |
| 6 | 2 N |
| 7a | 225,000 J |
| 7b | 225,000 J |
| 8 | 0 J |
| 9 | 57,600 J |
| 10a | 0 J |
| 10b | 5,998 J |
| 10c | 14,990 J |
| 11 | 980,000 J |

Let me know if you'd like explanations in simpler terms or diagrams!