Solveeit Logo
Login

Question

Question: A plank of mass M and length l is placed on a smooth surface and a man of mass ‘m’ is standing on th...

A plank of mass M and length l is placed on a smooth surface and a man of mass ‘m’ is standing on the plank at one end as shown in figure. Now the man starts running on a plank with constant acceleration ‘a’ with respect to the ground. The time taken by the man to reach other end of the plank is equal to

A

2lm⁡/a(m+M)\sqrt { 2 \operatorname { lm } / \mathrm { a } ( \mathrm { m } + \mathrm { M } ) }

B

lm⁡/a(m+M)\sqrt { \operatorname { lm } / \mathrm { a } ( \mathrm { m } + \mathrm { M } ) }

C

21/a\sqrt { 21 / a }

D

21M/am\sqrt { 21 \mathrm { M } / \mathrm { am } }

Answer

2lm⁡/a(m+M)\sqrt { 2 \operatorname { lm } / \mathrm { a } ( \mathrm { m } + \mathrm { M } ) }

Explanation

Solution

Plank exerts friction ‘f’ in forward direction on man. Appling N/L on the man w.r.t. ground

Same friction force is exerted by the man on the plank in backward direction

If acceleration of the plank is ‘b’ then

ma = Mb ⇒ b = ma/M

Net acc. of man w.r.t plank = a + b

= a + ma/M = a(M + m/M)

Using S = ut + 1/2 at2 for motion of the man with respect to the plank

l = 0 + 12a(m+MM)t2\frac { 1 } { 2 } \mathrm { a } \left( \frac { \mathrm { m } + \mathrm { M } } { \mathrm { M } } \right) \mathrm { t } ^ { 2 }

t = 2lm⁡a(m+M)\sqrt { \frac { 2 \operatorname { lm } } { \mathrm { a } ( \mathrm { m } + \mathrm { M } ) } }