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
- Total work done (W) = 300 J
- Mass of the block (m) = 2 kg
- Height of the inclined plane (h) = 10 m
- Acceleration due to gravity (g) = 10 m/s²

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

### Step 3: Set up the equation for total work done
The total work done is the sum of the work done against friction (Wf) and the work done against gravity (Wg):
\[ W = W_f + W_g \]
Substituting the known values:
\[ 300 \, \text{J} = W_f + 200 \, \text{J} \]

### Step 4: Solve for work done against friction
Rearranging the equation to find the work done against friction:
\[ W_f = 300 \, \text{J} - 200 \, \text{J} \]
\[ W_f = 100 \, \text{J} \]

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

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