A block of mass m = 1 kg is placed over a plank Q of mass M = 6 kg, placed over a smooth horizontal surface as shown in figure. Block P is given a velocity `v = 2 m//s^(2)` to the right. If the coefficient of friction between P and Q is `mu = 0.3`. Find the acceleration of Q relative to P.

Options

A. `4m//s^(2)`
B. `3.5m//s^(2)`
C. `2m//s^(2)`
D. `10.0m//s^(2)`

Answer

B

Explanation

Frictional force between P and Q is f = `mu m g` which will retard P and accelerate Q.
Retradation of P is `a_(p) =- (f)/(m) = (-mu mg)/(m) = - mu g`
Acceleration of Q is `a_(Q) = (+f)/(M) = (mu m g)/(M)`
Acceleration Q relative to P is
`a_(QP) =a_(Q)-a_(p)=(mumg)/(M)-(-mug)`
` = mu [1 + (m)/(M)]`
`=0.3 xx 10 [1 +(1)/(6)]`
`=3.5 m//s^(2)`