300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m . Work done agains

Question

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m . Work done against friction is (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )\

Professor PC Thomas Classes · Answer

(c) : Gain in potential energy $=m g h$ $$ =2 	imes 10 	imes 10=200 \mathrm{~J} $$ Gain in potential energy + work done against friction $$ =	ext { work done }=300 \mathrm{~J} $$ $	herefore$ Work done against friction $=300-200=100 \mathrm{~J}$\