Work Done by Frictional Force on Inclined Plane Moving Upward

To find the work done by the frictional force, we analyze the forces acting on the block and its displacement in the given time frame.

1. Given Data:

• Mass of the block, $m = 1\text{ kg}$
• Angle of inclination, $\theta = 30^\circ$
• Upward velocity of the assembly, $v = 4\text{ m/s}$
• Time duration, $t = 2\text{ s}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

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2. Analysis of Forces:

Since the assembly moves at a constant velocity, the acceleration of the block is zero ($\vec{a} = 0$). Thus, the net force on the block is zero.

The force of friction ($\vec{f}$) prevents the block from sliding down the inclined plane, so it acts along the inclined plane in the upward direction: $f = mg \sin\theta$

Substituting the given values: $f = 1\text{ kg} \times 10\text{ m/s}^2 \times \sin 30^\circ = 10 \times \frac{1}{2} = 5\text{ N}$

Since the inclined plane makes an angle of $\theta = 30^\circ$ with the horizontal, the component of the frictional force in the vertical direction ($f_y$) is: $f_y = f \sin\theta = 5 \times \sin 30^\circ = 5 \times \frac{1}{2} = 2.5\text{ N}$

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3. Displacement of the Assembly:

The assembly moves vertically upward at a constant speed of $v = 4\text{ m/s}$ for $t = 2\text{ s}$. The vertical displacement ($\vec{S}$) is: $S_y = v \times t = 4\text{ m/s} \times 2\text{ s} = 8\text{ m}$

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4. Work Done by the Frictional Force:

The work done by the frictional force ($W_f$) is the dot product of the friction force vector and the displacement vector: $W_f = \vec{f} \cdot \vec{S} = f_y \times S_y$

Substituting the values: $W_f = 2.5\text{ N} \times 8\text{ m} = 20\text{ J}$

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Conclusion:

The work done by the frictional force in time $2\text{ s}$ is $20\text{ J}$.

Correct Option: A