Work Practice Problems Worksheet #1 | Schemes and Mind Maps ... - Free Printable

Let's solve each of these Work Practice Problems step by step, using the formula for work:

🔷 Formula for Work:

$$
\text{Work} = \text{Force} \times \text{Displacement} \times \cos(\theta)
$$

In most of these cases, the force is applied in the direction of displacement (so $\cos(0^\circ) = 1$), so:
$$
W = F \times d
$$
Where:
- $ W $ = work in Joules (J)
- $ F $ = force in Newtons (N)
- $ d $ = displacement in meters (m)

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✔ Problem 1:

Amy uses 20 N of force to push a lawn mower 10 meters. How much work does she do?

$$
W = F \times d = 20\, \text{N} \times 10\, \text{m} = 200\, \text{J}
$$

✔ Answer: 200 Joules

---

✔ Problem 2:

How much work does an elephant do while moving a circus wagon 20 meters with a pulling force of 200 N?

$$
W = F \times d = 200\, \text{N} \times 20\, \text{m} = 4000\, \text{J}
$$

✔ Answer: 4000 Joules

---

✔ Problem 3:

Alex applies 350 N of force to move his stalled car 40 m. How much work did Alex do?

$$
W = F \times d = 350\, \text{N} \times 40\, \text{m} = 14,000\, \text{J}
$$

✔ Answer: 14,000 Joules

---

✔ Problem 4:

Tommy does 15 Joules of work to push the pencil over 1 meter. How much force did he use?

We rearrange the formula:
$$
F = \frac{W}{d} = \frac{15\, \text{J}}{1\, \text{m}} = 15\, \text{N}
$$

✔ Answer: 15 Newtons

---

✔ Problem 5:

Angela uses a force of 25 Newtons to lift her grocery bag while doing 50 Joules of work. How far did she lift the grocery bags?

$$
d = \frac{W}{F} = \frac{50\, \text{J}}{25\, \text{N}} = 2\, \text{m}
$$

✔ Answer: 2 meters

---

✔ Problem 6:

The baseball player does 1234 Joules of work when hitting a baseball into left field. Assuming the baseball landed 100 meters away from home plate, how much force did the player use to hit the ball?

Note: This assumes the entire 100 m is the displacement due to the force — but in reality, the force is only applied during contact (a very short distance). However, since the problem says "assuming the baseball landed 100 meters away", we’ll treat it as if the force was applied over that entire distance (which is not physically accurate, but acceptable for this problem).

$$
F = \frac{W}{d} = \frac{1234\, \text{J}}{100\, \text{m}} = 12.34\, \text{N}
$$

✔ Answer: 12.34 Newtons

> ⚠️ Note: In real life, the force is applied over a tiny distance (like 0.5 m), so the actual force would be much higher. But based on the assumption given, this is correct.

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## 📌 Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 200 J |
| 2 | 4000 J |
| 3 | 14,000 J |
| 4 | 15 N |
| 5 | 2 m |
| 6 | 12.34 N |

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## 🧩 Explain the following situation with diagram and labels:

You are given a diagram showing:
```
F → [block] → displacement
```

This means:
- A force F is applied to a block.
- The block moves in the same direction as the force.
- There is displacement in the direction of the force.

❓ Was there work done?

✔ Yes, because:
- A force was applied.
- The object moved (displaced).
- The displacement was in the direction of the force.

So:
$$
W = F \cdot d > 0
$$

🔬 Energy Description:

- Work is energy transferred by a force.
- The kinetic energy of the block increases as it accelerates due to the force.
- If the block started at rest, the work done equals the change in kinetic energy.
- The source of energy could be the person pushing or the system applying the force.

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🖼️ Diagram Explanation (Text-based):

```
→ F
[ ] → displacement
Block
```

- F: Applied force (vector pointing right)
- Block: Object being pushed
- → displacement: Direction of movement (same as force)

✔️ Conclusion:

- Work was done because force and displacement are in the same direction.
- Energy transfer occurred: Mechanical energy (kinetic) increased.
- Work = Force × Displacement → Positive value.

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Let me know if you'd like a hand-drawn version of the diagram or further clarification!