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. 50 J
B. 100 J
C. 200 J
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 Given Values:**
- Mass of the block (m) = 5 kg
- Height of the incline (h) = 4 m
- Total work done (Wt) = 250 J
- Acceleration due to gravity (g) = 10 m/s²

2. **Calculate Work Done Against Gravity (Wg):**
The work done against gravity can be calculated using the formula:
\[ Wg = m \cdot g \cdot h \]
Substituting the known values:
\[ Wg = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 4 \, \text{m} = 200 \, \text{J} \]

3. **Set Up the Equation for Total Work Done:**
The total work done (Wt) is the sum of the work done against friction (Wf) and the work done against gravity (Wg):
\[ Wt = Wf + Wg \]
Rearranging this gives us:
\[ Wf = Wt - Wg \]

4. **Substitute the Values:**
Now, substitute the values of Wt and Wg into the equation:
\[ Wf = 250 \, \text{J} - 200 \, \text{J} = 50 \, \text{J} \]

5. **Conclusion:**
The work done against friction (Wf) is 50 J.

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

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