If 250 J of work is done in sliding a 5 kg block up an inclined plane of height 4 m. Work done against friction is `(g=10ms^(-2))`

Options

A. `50J`
B. `100J`
C. `200J`
D. Zero

Answer

To solve the problem, we need to find the work done against friction when sliding a 5 kg block up an inclined plane to a height of 4 m, given that the total work done is 250 J.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Total work done (W_total) = 250 J
- Mass of the block (m) = 5 kg
- Height of the incline (h) = 4 m
- Acceleration due to gravity (g) = 10 m/s²

2. **Calculate the Weight of the Block:**
\[ \text{Weight (W)} = m \cdot g = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \]

3. **Calculate the Work Done Against Gravity:**
\[ \text{Work done against gravity (W_gravity)} = \text{Weight} \cdot \text{Height} = W \cdot h = 50 \, \text{N} \cdot 4 \, \text{m} = 200 \, \text{J} \]

4. **Use the Work-Energy Principle:**
The total work done is the sum of the work done against gravity and the work done against friction.
\[ W_{\text{total}} = W_{\text{friction}} + W_{\text{gravity}} \]

5. **Rearranging to Find Work Done Against Friction:**
\[ W_{\text{friction}} = W_{\text{total}} - W_{\text{gravity}} = 250 \, \text{J} - 200 \, \text{J} = 50 \, \text{J} \]

6. **Conclusion:**
The work done against friction is 50 J.

### Final Answer:
The work done against friction is **50 J**.

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