300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction is

Options

A. 1000 J
B. 200 J
C. 100 J
D. Zero

Answer

C

Explanation

P.E of block 2 kg at height , h = 10 m
P.E = mgh `= 2xx 10 xx 10 = 200 J `
Total work done due to sliding on inclined plane = 300 J
Work done against friction force ` = (300 - 200)`
W = 100 J