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Question: A body is placed in the middle of a plank of length \( l \) . The coefficient of friction between th...

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

Explanation

Solution

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+12at2s = ut + \dfrac{1}{2}a{t^2} where ss is the distance travelled by a body, uu is the initial velocity, tt is time taken to travel the distance, and aa is the acceleration of the body.
F=maF = ma where FNET{F_{NET}} is the net force acting on a body, mm is the mass of the body, and aa 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 μ\mu .
To find this, we recall the distance travelled by the body from the equation of motion
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2} where ss is the distance travelled by a body, uu is the initial velocity, tt is time taken to travel the distance, and aa is the acceleration of the body.
Since, the body was at rest u=0u = 0 , then
s=12at2s = \dfrac{1}{2}a{t^2}
Hence, by multiplying both sides by 2 and dividing by aa , then square rooting, the time taken can be given as
t=2sat = \sqrt {\dfrac{{2s}}{a}}
Since friction acts on the body, this reduces the acceleration by an amount μg\mu g . And also the distance is s=l2s = \dfrac{l}{2}
Hence,
t=2l2aμg=laμgt = \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 μg\mu g can be proven as follows. From Newton’s second law
FNET=manet{F_{NET}} = m{a_{net}}
Now, whatever force first accelerated the body, it can be quantified as
F=maF = ma where ai{a_i} is the initial acceleration when the force acted on the body.
But net force would be given as
Ff=manetF - f = m{a_{net}} where ff is friction, then
maiμmg=manetm{a_i} - \mu mg = m{a_{net}} (since frictional force is given as μmg\mu mg )
Cancelling the mass, then final acceleration is given by anet=aiμg{a_{net}} = {a_i} - \mu g .