Question

Friction is sufficient so that the block does not slide. Find the work done by friction in \(2\) sec. 

• A. \(45\)
• B. \(40\)
• C. \(20\)
• D. \(10\)

Hint:

When a block does not slide on an accelerating or moving surface, friction adjusts to maintain relative equilibrium. Always resolve friction along the direction of displacement to calculate work.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The wedge is moving vertically upwards with a constant velocity of 4 m/s.
Since the block does not slide relative to the wedge, it also moves upwards with a constant velocity of 4 m/s.
The work done by friction is the dot product of the frictional force and the displacement of the block.
Step 2: Key Formula or Approach:
Work done is given by $W = \vec{f}_r \cdot \vec{S}$.
This can be simplified to $W = f_{ry} \times S_y$, where $f_{ry}$ is the vertical component of the frictional force and $S_y$ is the vertical displacement.
Step 3: Detailed Explanation:
Since the block is moving with a constant velocity, the net force on it is zero.
Along the incline, the downward force due to gravity is $mg \sin 30^\circ$.
For the block to remain stationary on the wedge, static friction $f_r$ must act up the incline to balance this force.
\[ f_r = mg \sin 30^\circ = 1 \times 10 \times \frac{1}{2} = 5 \text{ N} \]
The direction of friction is up the incline, making an angle of $30^\circ$ with the horizontal, which is $60^\circ$ with the vertical axis.
Thus, the vertical component of the frictional force is:
\[ (f_r)_y = f_r \sin 30^\circ = 5 \times \frac{1}{2} = 2.5 \text{ N} \]
The vertical displacement of the block in $t = 2 \text{ sec}$ is:
\[ S_y = v \times t = 4 \text{ m/s} \times 2 \text{ s} = 8 \text{ m} \]
The work done by friction is:
\[ W = (f_r)_y \times S_y = 2.5 \text{ N} \times 8 \text{ m} = 20 \text{ J} \]
Step 4: Final Answer:
The work done by friction in 2 sec is 20.