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Question: A block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N ...

A block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N on a rough horizontal surface. The coefficient of friction between block and surface is 0.1. The work done by friction in 10 s is (Take g=10 ms−2\mathrm { g } = 10 \mathrm {~ms} ^ { - 2 })

A

200 J

B

-200 J

C

600 J

D

-600 J

Answer

-200 J

Explanation

Solution

The frictional force and the displacement are in opposite direction, therefore the angel between frictional force and displacement is θ=180∘\theta = 180 ^ { \circ }

Thus,

Work done by the friction is

Wf=fdcos⁡180∘=(2 N)(100 m)cos⁡180∘=−200 J\mathrm { W } _ { \mathrm { f } } = \mathrm { fd } \cos 180 ^ { \circ } = ( 2 \mathrm {~N} ) ( 100 \mathrm {~m} ) \cos 180 ^ { \circ } = - 200 \mathrm {~J}