JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2026
- Medium
- MCQ
A mass of \(1\text{ kg}\) is kept on an inclined plane with \(30°\) inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of \(4\text{ m/s}\). The work done by the frictional force in time \(2\text{ s}\) is _______ J. (Take \(g = 10\text{ m/s}^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
Since the assembly moves with a constant velocity, the acceleration of the mass is zero. The mass is in equilibrium relative to the incline.
The frictional force \(f\) acting on the mass balances the component of gravity along the inclined plane. Therefore, the magnitude of the frictional force is:
\(f = mg \sin 30^\circ\)
Substituting the given values (\(m = 1\text{ kg}\), \(g = 10\text{ m/s}^2\)):
\(f = 1 \times 10 \times \dfrac{1}{2} = 5\text{ N}\)
The direction of the frictional force is upwards along the incline. Since the incline makes an angle of \(30^\circ\) with the horizontal, the frictional force makes an angle of \(90^\circ - 30^\circ = 60^\circ\) with the vertical upward direction.
The displacement \(s\) of the assembly in time \(t = 2\text{ s}\) is vertically upwards:
\(s = vt = 4 \times 2 = 8\text{ m}\)
The work done by the frictional force is:
\(W = f s \cos 60^\circ\)
\(W = 5 \times 8 \times \dfrac{1}{2} = 20\text{ J}\)
Answer: \(20\)
The frictional force \(f\) acting on the mass balances the component of gravity along the inclined plane. Therefore, the magnitude of the frictional force is:
\(f = mg \sin 30^\circ\)
Substituting the given values (\(m = 1\text{ kg}\), \(g = 10\text{ m/s}^2\)):
\(f = 1 \times 10 \times \dfrac{1}{2} = 5\text{ N}\)
The direction of the frictional force is upwards along the incline. Since the incline makes an angle of \(30^\circ\) with the horizontal, the frictional force makes an angle of \(90^\circ - 30^\circ = 60^\circ\) with the vertical upward direction.
The displacement \(s\) of the assembly in time \(t = 2\text{ s}\) is vertically upwards:
\(s = vt = 4 \times 2 = 8\text{ m}\)
The work done by the frictional force is:
\(W = f s \cos 60^\circ\)
\(W = 5 \times 8 \times \dfrac{1}{2} = 20\text{ J}\)
Answer: \(20\)
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