A mass of 1 text{ kg} is kept on an inclined | vedclass

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(A) The mass is on an inclined plane with an angle of inclination $	heta = 30^\circ$. Since the assembly moves with a constant velocity, the net acce

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$20$

Vedclass · Answer

(A) The mass is on an inclined plane with an angle of inclination $	heta = 30^\circ$. Since the assembly moves with a constant velocity, the net acceleration of the block is zero. The frictional force $f$ acting on the block to keep it at rest relative to the inclined plane is $f = mg \sin 	heta$. Substituting the given values: $f = 1 	imes 10 	imes \sin(30^\circ) = 1 	imes 10 	imes 0.5 = 5 	ext{ N}$. The assembly moves with a constant velocity $v = 4 	ext{ m/s}$ for a time $t = 2 	ext{ s}$. The displacement $s$ of the block in the direction of motion is $s = v 	imes t = 4 	imes 2 = 8 	ext{ m}$. The work done by the frictional force is the product of the force and the displacement in the direction of the force. Since the frictional force acts along the incline and the displacement is also along the incline, the work done is $W = f 	imes s = 5 	imes 8 = 40 	ext{ J}$. However, considering the work done by the frictional force relative to the ground, the force is $5 	ext{ N}$ and the displacement is $8 	ext{ m}$, resulting in $40 	ext{ J}$. Given the options provided, there appears to be a discrepancy in the expected answer format or context, but based on standard physics principles, the work done is $40 	ext{ J}$. If we re-evaluate the component of displacement along the incline, the result remains $40 	ext{ J}$.

Column-$I$	Column-$II$
$(1)$ Static friction	$(a)$ Limiting friction
$(2)$ Rolling friction	$(b)$ Ball bearing
	$(c)$ Object moving on a road

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Match Column-$I$ with Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Static friction $(a)$ Limiting friction

$(2)$ Rolling friction $(b)$ Ball bearing

$(c)$ Object moving on a road

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