In the arrangement shown, on external agent slowly pulls the block P from A to B along the rough inclined surface AB. The coefficient of friction between the block and the wedge is 0.75. Then the ratio of changes in potential energy of the block to the work done by the agent is

Question

Infinity Learn · Accepted Answer

(3)Δ$$U=Change$$ in  potential  energy  $$=mghwf=work$$  done  against  fractional  $$force=f$$× $$AB=$$μmgcosθ  ×  hsinθ$$=$$μ$$mgh$$. $$cot$$θ∴  $$w=work$$  done by  external  agent $$=wf+$$Δ$$U=mgh1+$$μ$$cos$$θ∴  Δ$$Uw=mghmgh1+$$μ$$cot$$θ$$=$$  $$11+34cot$$ $$450$$ $$=47$$

In the arrangement shown, on external agent slowly pulls the block P from A to B along the rough inclined surface AB. The coefficient of friction between the block and the wedge is 0.75. Then the ratio of changes in potential energy of the block to the work done by the agent is

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