Work Power and Energy 3 Question 10

10. In the figures (a) and (b) AC,DG and GF are fixed inclined planes, BC=EF=x and AB=DE=y. A small block of mass M is released from the point A. It slides down AC and reaches C with a speed vC. The same block is released from rest from the point D. It slides down DGF and reaches the point F with speed vF. The coefficients of kinetic frictions between the block and both the surfaces AC and DGF are μ. Calculate vC and vF.

(1980,6M)

(a)

(b)

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Answer:

Correct Answer: 10. vC=vF=2(gyμgx)

Solution:

  1. In both the cases, work done by friction will be μMgx.

12MvC2=12MvF2=MgyμMgxvC=vF=2gy2μgx



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