300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction is

Options

A. 200J
B. 100J
C. zero
D. 1000J

Answer

To solve the problem, we need to determine the work done against friction when sliding a block up an inclined plane. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Work done (W) = 300 J
- Mass of the block (m) = 2 kg
- Height of the incline (h) = 10 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the work done against gravity
The work done against gravity (W_gravity) can be calculated using the formula:
\[ W_{\text{gravity}} = m \cdot g \cdot h \]
Substituting the values:
\[ W_{\text{gravity}} = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 10 \, \text{m} \]
\[ W_{\text{gravity}} = 200 \, \text{J} \]

### Step 3: Use the work-energy principle
According to the work-energy principle, the total work done (W) is equal to the work done against friction (W_friction) plus the work done against gravity (W_gravity):
\[ W = W_{\text{friction}} + W_{\text{gravity}} \]

### Step 4: Rearrange the equation to find work done against friction
Rearranging the equation gives us:
\[ W_{\text{friction}} = W - W_{\text{gravity}} \]
Substituting the known values:
\[ W_{\text{friction}} = 300 \, \text{J} - 200 \, \text{J} \]
\[ W_{\text{friction}} = 100 \, \text{J} \]

### Conclusion
The work done against friction is **100 J**.

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