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 kg
- Height of the slant plane (h) = 10 m
- Total work done (W_total) = 300 J

2. **Calculate the Gravitational Potential Energy (GPE):**
The work done against gravity when lifting the block to a height \( h \) is given by the formula:
\[ \text{GPE} = m \cdot g \cdot h \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Substituting the values:
\[ \text{GPE} = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{m} = 196.2 \, \text{J} \]

3. **Determine the Work Done Against Friction:**
The total work done (W_total) is the sum of the work done against gravity (GPE) and the work done against friction (W_friction):
\[ W_{\text{total}} = \text{GPE} + W_{\text{friction}} \]
Rearranging this gives:
\[ W_{\text{friction}} = W_{\text{total}} - \text{GPE} \]

Substituting the values we have:
\[ W_{\text{friction}} = 300 \, \text{J} - 196.2 \, \text{J} = 103.8 \, \text{J} \]

4. **Conclusion:**
The work done against friction is \( 103.8 \, \text{J} \).

### Final Answer:
The work done against friction is **103.8 J**.