A box of mass $$50kg$$ at rest is pulled up on an inclined plane $$12m$$ long and $$2m$$ high by a constant force of $$100N$$ applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is $$2ms$$–$$1$$, the work done against friction in Joules is (g$$ $$=$$ $$10ms$$–$$2)

Question

Infinity Learn · Accepted Answer

WF $$=$$ Work done by the applied force $$=$$ $$10$$ x $$12$$ $$=1200J$$ΔV $$=$$ Change in P.E $$=$$ $$mgh$$ $$=$$ $$50$$ x $$10$$ x $$2$$ $$=$$ $$1000J$$ΔK $$=$$ Change in K.E. $$=$$ $$1/2$$ x $$50$$ x $$22$$ $$=$$ $$100JNow$$, WF $$=$$ ΔK $$+$$ ΔV $$+$$ lost energy Lost energy $$=$$ $$(1200-100-1000)J$$ $$=$$ $$100J$$.

A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms^–1, the work done against friction in Joules is (g = 10ms^–2)

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A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms–1, the work done against friction in Joules is (g = 10ms–2)

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A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms^–1, the work done against friction in Joules is (g = 10ms^–2)