300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g=10 m/s2, work done against friction is

Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g=10 m/s2, work done against friction is (A) 200 J (B) 100 J (C) zero (D) 1000 J

Tardigrade Admin · Accepted Answer

Net work done in sliding a body up to a height

h

on inclined plane

=

Work done against gravitational force

+

Work done against frictional force

\Rightarrow W = W_{g} + W_{f} \dots

(i)
but

W = 300 J

W_{g} = m g h = 2 \times 10 \times 10 = 200 J

Putting in Eq. (i), we get

300 = 200 + W_{f}

\Rightarrow W_{f} = 300 - 200 = 100 J

Q. 300J of work is done in sliding a 2kg block up an inclined plane of height 10m. Taking g=10m/s2, work done against friction is

200 J

100 J

zero

1000 J

Net work done in sliding a body up to a height h on inclined plane = Work done against gravitational force + Work done against frictional force ⇒W=Wg​+Wf​…(i) but W=300J Wg​=mgh=2×10×10=200J Putting in Eq. (i), we get 300=200+Wf​ ⇒Wf​=300−200=100J

Q. $300 J$ of work is done in sliding a $2 k g$ block up an inclined plane of height $10 m$ . Taking $g = 10 m / s^{2},$ work done against friction is