Question

A mass of 1 kg is kept on a inclined plane with $30^\circ$ inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2 s is \dots J. (Take $g = 10 \text{ m/s}^2$)

• A. 20
• B. 25
• C. 30
• D. 10

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Work done is $W = \vec{f} \cdot \vec{d}$. Since the assembly moves with constant velocity, the net force on the block is zero. Friction must balance the component of gravity.
Step 2: Detailed Explanation:
Mass $m = 1 kg$, angle $\theta = 30^\circ$.
The block is at rest relative to the incline.
Static friction $f = mg \sin \theta$ (acting up the incline).
$f = 1 \cdot 10 \cdot \sin 30^\circ = 5 N$.
The assembly moves vertically up with $v = 4 m/s$.
Displacement in 2 s is $d = v \cdot t = 4 \cdot 2 = 8 m$ (upwards).
Angle between friction force (along incline at $30^\circ$ to horizontal) and vertical displacement is $60^\circ$.
Work done $W = f \cdot d \cdot \cos 60^\circ$
$W = 5 \cdot 8 \cdot \frac{1}{2} = 20 J$.
Step 3: Final Answer:
The work done by friction is 20 J.