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Question

A 5  kg block slides down an inclined  plane, having angle of inclination 30 from horizontal. Find the acceleration  (m/s2) of block, if the coefficient of kinetic friction is μk=123. Take g=10 m/s2 .
  1. 3.5 m/s2
  2. 10 m/s2
  3. 7.5 m/s2
  4. 2.5 m/s2


Solution

The correct option is D 2.5 m/s2

Since the block is sliding down the inclined plane with acceleration a, kinetic friction will act:
fk=μkN
From the equilibrium condition for  block, to inclined plane:
N=mgcosθ.......i

Applying newton's 2nd law along the inclined plane in direction of acceleration:
mgsinθμkN=ma...........ii

From Eq i and ii:

mgsin30μkmgcos30=ma

a=gsin30μkgcos30
a=g21233g2
a=52=2.5 m/s2

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