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=12√3. Take g=10 m/s2 .3.5 m/s210 m/s27.5 m/s22.5 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: mgsinθ−μkN=ma...........ii From Eq i and ii: mgsin30−μkmgcos30=ma ⇒a=gsin30−μkgcos30∘ ∴a=g2−12√3√3g2 ∴a=52=2.5 m/s2

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