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 250 J of work is done in total.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of the block (m) = 5 kg
- Height of the incline (h) = 4 m
- Work done (W_total) = 250 J
- 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 (potential energy gained):**
\[ W_{\text{gravity}} = m \cdot g \cdot h = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 4 \, \text{m} = 200 \, \text{J} \]

4. **Use the work-energy principle:**
The total work done is the work done against gravity plus the work done against friction.
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{friction}} \]

5. **Rearranging the equation 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**.