A block of mass 2kg slipped up a slant plane requires 300J of work. If height of slant is 10m the work done against friction is :

Options

A. 100 J
B. 200 J
C. 300 J
D. zero

Answer

To solve the problem, we need to determine the work done against friction when a block of mass 2 kg is moved up a slant plane to a height of 10 m, requiring a total work of 300 J.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the block, \( m = 2 \, \text{kg} \)
- Height of the slant plane, \( h = 10 \, \text{m} \)
- Total work done, \( W_{\text{total}} = 300 \, \text{J} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (assuming standard value)

2. **Calculate the Work Done Against Gravity:**
The work done against gravity (potential energy gained) when lifting the block to height \( h \) is given by the formula:
\[ W_{\text{gravity}} = mgh \]
Substituting the values:
\[ W_{\text{gravity}} = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 200 \, \text{J} \]

3. **Set Up the Equation for Total Work:**
The total work done is the sum of the work done against gravity and the work done against friction:
\[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{friction}} \]
Substituting the known values:
\[ 300 \, \text{J} = 200 \, \text{J} + W_{\text{friction}} \]

4. **Solve for Work Done Against Friction:**
Rearranging the equation to find \( W_{\text{friction}} \):
\[ W_{\text{friction}} = 300 \, \text{J} - 200 \, \text{J} = 100 \, \text{J} \]

5. **Conclusion:**
The work done against friction is \( 100 \, \text{J} \).

### Final Answer:
The work done against friction is **100 J**.