To solve the problem, we need to find the work done by the force of friction on a block of mass `1 kg` placed on a rough inclined wedge of inclination `45 degrees`, while the wedge is in an elevator moving upwards with a uniform velocity of `2 m/s`.
### Step-by-Step Solution:
1. **Identify the Forces Acting on the Block:**
- The weight of the block, \( W = mg \), acts downwards.
- The normal force \( N \) acts perpendicular to the surface of the wedge.
- The frictional force \( f \) acts parallel to the surface of the wedge, opposing the component of the weight that tends to slide the block down.
2. **Calculate the Weight of the Block:**
- Given \( m = 1 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \),
\[
W = mg = 1 \times 10 = 10 \, \text{N}
\]
3. **Resolve the Weight into Components:**
- The component of weight parallel to the incline:
\[
W_{\parallel} = mg \sin \theta = 10 \sin 45^\circ = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{N}
\]
- The component of weight perpendicular to the incline:
\[
W_{\perpendicular} = mg \cos \theta = 10 \cos 45^\circ = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{N}
\]
4. **Determine the Normal Force:**
- Since the block does not slide, the normal force \( N \) balances the perpendicular component of the weight:
\[
N = W_{\perpendicular} = \frac{10}{\sqrt{2}} \, \text{N}
\]
5. **Calculate the Frictional Force:**
- The frictional force \( f \) must equal the component of weight down the incline:
\[
f = W_{\parallel} = \frac{10}{\sqrt{2}} \, \text{N}
\]
6. **Calculate the Displacement of the Block in 1 Second:**
- The elevator moves up with a velocity of \( 2 \, \text{m/s} \). In 1 second, the displacement \( S \) is:
\[
S = v \times t = 2 \times 1 = 2 \, \text{m}
\]
7. **Calculate the Work Done by the Friction Force:**
- The work done by the friction force \( W_f \) is given by:
\[
W_f = f \times S \times \cos(\theta)
\]
- Here, \( \theta = 45^\circ \), so \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[
W_f = \left(\frac{10}{\sqrt{2}}\right) \times 2 \times \frac{1}{\sqrt{2}} = \frac{10 \times 2}{2} = 10 \, \text{J}
\]
### Final Answer:
The work done by the force of friction on the block in 1 second is **10 joules**.
