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

Tardigrade Admin · Accepted Answer

Loss in potential energy =m g h =2 × 10 × 10 =200 J Gain in kinetic energy = work done =300 J ∴ Work done against friction =300-200=100 J

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

1000 J

200 J

100 J

zero

Loss in potential energy $= m g h$
$= 2 \times 10 \times 10$
$= 200 J$
Gain in kinetic energy $=$ work done $= 300 J$
$∴$ Work done against friction $= 300 - 200 = 100 J$

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