Question

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 the applied force in 10 s is (Take $\mathrm { g } = 10 \mathrm {~ms} ^ { - 2 }$)

• A. 200 J
• B. -200 J
• C. 600 J
• D. -600 J

Answer 1

Answer: (C)

600 J

Answer 2

The various forces acting on the block is as Shown in the figure.

Here, $\mathrm { m } = 2 \mathrm {~kg} , \mu = 0.1 , \mathrm {~F} = 6 \mathrm {~N} , \mathrm {~g} = 10 \mathrm {~ms} ^ { - 2 }$

Force of friction,

$\mathrm { f } = \mu \mathrm { N } = 0.1 \times 2 \mathrm {~kg} \times 10 \mathrm {~ms} ^ { - 2 } = 2 \mathrm {~N}$

Net force with which the block moves

Net acceleration with which the block moves

Distance travelled by the block in 10 s is

$\mathrm { d } = \frac { 1 } { 2 } \mathrm { at } ^ { 2 } = \frac { 1 } { 2 } \times 2 \mathrm {~ms} ^ { - 2 } ( 10 \mathrm {~s} ) ^ { 2 } = 100 \mathrm {~m} \quad ( \therefore \mathrm { u } = 0 )$

As the applied force and displacement are in the same direction, therefore angle between the applied force and the displacement is $\theta = 0 ^ { 0 }$

Hence,

Work done by the applied force,