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A 5.0 kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.5. A horizontal force pulls the box at constant velocity for 10 cm. The work done by the applied horizontal force and the frictional force are respectively (take g=10m/s2)

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a

2.5 J and 2.5 J

b

zero and 2.5 J

c

2.5 J and zero

d

2.5 J and –2.5J

answer is D.

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Detailed Solution

As velocity is constant work done by the applied force is only to overcome the frictional force.so work done by the applied force is equal to μmgs=0.5×5×10×0.1=2.5J by frictional force=-2.5 J

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A 5.0 kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.5. A horizontal force pulls the box at constant velocity for 10 cm. The work done by the applied horizontal force and the frictional force are respectively (take g=10m/s2)