Q.

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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a

4 : 5

b

2 : 7

c

4 : 7

d

3 : 5

answer is C.

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Detailed Solution

(3)ΔU=Change in  potential  energy  =mghwf=work  done  against  fractional  force=f× AB=μmgcosθ  ×  hsinθ=μmgh. cotθ∴  w=work  done by  external  agent =wf+ΔU=mgh1+μcosθ∴  ΔUw=mghmgh1+μcotθ=  11+34cot 450 =47
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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