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

Answer

B

Explanation

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