Physics worksheet focusing on work and energy calculations with nine practice problems and solutions.

Let's solve each of these Work and Energy problems step by step, using the relevant physics formulas.

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1) A force of 10N causes a displacement of 2m in a body in its own direction. Calculate the work done by force. (20J)

Formula:
$$
\text{Work} = \text{Force} \times \text{Displacement} \times \cos\theta
$$
Since the force is in the direction of displacement, $\theta = 0^\circ$, so $\cos 0^\circ = 1$.

$$
W = F \cdot d = 10\,\text{N} \times 2\,\text{m} = 20\,\text{J}
$$

✔ Answer: 20 J

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2) A body of mass 2kg is moving with a speed of 20m/s. Find the kinetic energy. (400J)

Formula:
$$
KE = \frac{1}{2}mv^2
$$

$$
KE = \frac{1}{2} \times 2\,\text{kg} \times (20\,\text{m/s})^2 = 1 \times 400 = 400\,\text{J}
$$

✔ Answer: 400 J

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3) A moving body of 30kg has 60 J of KE. Calculate the speed.

Use:
$$
KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m}}
$$

$$
v = \sqrt{\frac{2 \times 60}{30}} = \sqrt{\frac{120}{30}} = \sqrt{4} = 2\,\text{m/s}
$$

✔ Answer: 2 m/s

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4) How much force is applied on the body when 150 joule of work is done in displacing the body through a distance of 10m in the direction of force? (15 N)

$$
W = F \cdot d \Rightarrow F = \frac{W}{d} = \frac{150\,\text{J}}{10\,\text{m}} = 15\,\text{N}
$$

✔ Answer: 15 N

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5) A body of 5kg raised to 2m. Find the work done. (98J)

When lifting a body vertically, work done against gravity:

$$
W = mgh
$$
Assume $g = 9.8\,\text{m/s}^2$

$$
W = 5\,\text{kg} \times 9.8\,\text{m/s}^2 \times 2\,\text{m} = 98\,\text{J}
$$

✔ Answer: 98 J

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6) A work of 4900 J is done on a load of mass 50 kg to lift it to a certain height. Calculate the height through which the load is lifted. (10m)

Again, use:
$$
W = mgh \Rightarrow h = \frac{W}{mg}
$$

$$
h = \frac{4900}{50 \times 9.8} = \frac{4900}{490} = 10\,\text{m}
$$

✔ Answer: 10 m

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7) An engine does 54,000 J of work by exerting a force of 6000 N on it. What is the displacement of the force? (10m)

$$
W = F \cdot d \Rightarrow d = \frac{W}{F} = \frac{54000}{6000} = 9\,\text{m}
$$

Wait! But the answer given is 10m, but our calculation gives 9m.

Let’s check:
$ 6000 \times 9 = 54,000 $ → Correct.

So either there's a typo in the problem or the expected answer is wrong.

But if we assume the answer is 10m, then:
$$
W = F \cdot d = 6000 \times 10 = 60,000\,\text{J} \neq 54,000\,\text{J}
$$

So the correct answer should be:

✔ Answer: 9 m

But since the sheet says (10m), double-check the values. Maybe it's 6000N and 54,000J, then:

$$
d = \frac{54000}{6000} = 9\,\text{m}
$$

✘ So the given answer (10m) is incorrect.

✔ Correct Answer: 9 m

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8) A machine does 192 J of work in 240 sec. What is the power of the machine? (8w)

$$
\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{192\,\text{J}}{240\,\text{s}} = 0.8\,\text{W}
$$

But the expected answer is 8W, which suggests a mistake.

Wait: $ 192 / 240 = 0.8 $, not 8.

But if the work was 1920 J, then $ 1920 / 240 = 8 $ W.

So likely, either:

- Work is 1920 J, or
- Time is 24 s, or
- The answer is 0.8 W

But as per given: 192 J in 240 s

$$
P = \frac{192}{240} = 0.8\,\text{W}
$$

✔ Correct Answer: 0.8 W

But the sheet says (8w) — this appears to be an error.

Alternatively, maybe the time is 24 seconds?

Then: $ 192 / 24 = 8\,\text{W} $

So probably a typo in the problem statement: 240 sec vs 24 sec

Assuming the intended time is 24 s, then:

✔ Answer: 8 W

But based on 240 sec, the correct answer is 0.8 W

Let’s go with what's written: 240 sec

✘ So the given answer (8W) is incorrect.

✔ Correct Answer: 0.8 W

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9) A force of 5N acting on a body at an angle of 30° with the horizontal direction displaces it horizontally through a distance of 6 m. Calculate the work done. (15√3 J)

Note: Angle is 30° with horizontal, and displacement is horizontal.

So only the horizontal component of force does work.

$$
F_{\text{horizontal}} = F \cos\theta = 5 \cos(30^\circ)
$$

$$
\cos(30^\circ) = \frac{\sqrt{3}}{2}
\Rightarrow F_{\text{horizontal}} = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}\,\text{N}
$$

Now, work done:
$$
W = F_{\text{horizontal}} \times d = \left(\frac{5\sqrt{3}}{2}\right) \times 6 = 15\sqrt{3}\,\text{J}
$$

✔ Answer: $15\sqrt{3}$ J

This matches the given answer.

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## ✔ Final Answers Summary:

| Q | Answer | Notes |
|---|--------|-------|
| 1 | 20 J | Correct |
| 2 | 400 J | Correct |
| 3 | 2 m/s | Correct |
| 4 | 15 N | Correct |
| 5 | 98 J | Correct |
| 6 | 10 m | Correct |
| 7 | 9 m | Given answer (10m) is wrong |
| 8 | 0.8 W | Given answer (8W) is wrong unless time is 24s |
| 9 | $15\sqrt{3}$ J | Correct |

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🛠️ Key Formulas Used:

- Work: $ W = F \cdot d \cdot \cos\theta $
- Kinetic Energy: $ KE = \frac{1}{2}mv^2 $
- Gravitational Potential Energy: $ PE = mgh $
- Power: $ P = \frac{W}{t} $

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