Work done by the friction force on $2 k g$ block upto $5$ sec with respect to $3 k g$ block is:

100 J

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400 J

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zero

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160 J

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Solution

The correct option is D zero
At $t = 5 s$ , External force will be $F = 25 N$ . Assume that both the blocks are still moving together, let us calculate the friction required between two blocks.

Friction from ground will be $0.02 \times 5 \times 50 = 5 N$ .
Hence, acceleration of the system will be $a = \frac{25 - 5}{5} = {4 m / s}^{2}$ .

For this acceleration, force required on above block is $F_{1} = 2 \times 4 = 8 N$ and maximum possible friction on above block is $0.4 \times 20 = 8 N$ . Since this condition is possible, both blocks will move together till $t = 5 s$ and hence work done by friction between two blocks will be 0.

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Q. In the figure shown below, coefficient of friction between the

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[Take

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A block of mass 2.0 kg is pushed down an inclined plane of inclination $37^{\circ}$ with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of $10 m / s^{2}$ . If the block started from rest, find the work done $(W_{F})$ by the applied force, work done $(W_{g})$ by the weight of the block and work done $(W_{f})$ by the frictional force acting on the block in the 1st second. Take $g = 10 m / s^{2}$ .