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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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