Question

A body is placed in the middle of a plank of length $ l $ . The coefficient of friction between the body and the plank is $ \mu $ . If the body starts moving with an acceleration $ a $ , what is the time after which the body leaves the plank.
(A) $ \sqrt {\dfrac{l}{{a - \mu g}}} $
(B) $ \sqrt {\dfrac{l}{{a + \mu g}}} $
(C) $ \sqrt {\dfrac{{2l}}{{a - \mu g}}} $
(D) $ \sqrt {\dfrac{{2l}}{{a + \mu g}}} $

Answer 1

Hint : Since the body was at the middle of the plank, the distance travelled before leaving the plank is half of the planck length. The frictional force will reduce the acceleration the body started moving with, giving it a lower net acceleration.

Formula used: In this solution we will be using the following formula;
 $ s = ut + \dfrac{1}{2}a{t^2} $ where $ s $ is the distance travelled by a body, $ u $ is the initial velocity, $ t $ is time taken to travel the distance, and $ a $ is the acceleration of the body.
 $ F = ma $ where $ {F_{NET}} $ is the net force acting on a body, $ m $ is the mass of the body, and $ a $ is the acceleration.

Complete step by step answer
A body was placed in the middle of a plank and was accelerated. The time taken for the body to leave the plank when the coefficient of friction between the plank and body is $ \mu $ .
To find this, we recall the distance travelled by the body from the equation of motion
 $ s = ut + \dfrac{1}{2}a{t^2} $ where $ s $ is the distance travelled by a body, $ u $ is the initial velocity, $ t $ is time taken to travel the distance, and $ a $ is the acceleration of the body.
Since, the body was at rest $ u = 0 $ , then
 $ s = \dfrac{1}{2}a{t^2} $
Hence, by multiplying both sides by 2 and dividing by $ a $ , then square rooting, the time taken can be given as
 $ t = \sqrt {\dfrac{{2s}}{a}} $
Since friction acts on the body, this reduces the acceleration by an amount $ \mu g $ . And also the distance is $ s = \dfrac{l}{2} $
Hence,
 $ t = \sqrt {\dfrac{{2\dfrac{l}{2}}}{{a - \mu g}}} = \sqrt {\dfrac{l}{{a - \mu g}}} $
Hence, the correct option is A.

Note
For clarity, the acceleration reduced by $ \mu g $ can be proven as follows. From Newton’s second law
 $ {F_{NET}} = m{a_{net}} $
Now, whatever force first accelerated the body, it can be quantified as
 $ F = ma $ where $ {a_i} $ is the initial acceleration when the force acted on the body.
But net force would be given as
 $ F - f = m{a_{net}} $ where $ f $ is friction, then
 $ m{a_i} - \mu mg = m{a_{net}} $ (since frictional force is given as $ \mu mg $ )
Cancelling the mass, then final acceleration is given by $ {a_{net}} = {a_i} - \mu g $ .