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Question: A man of mass \( 50kg \) is standing on a \( 100kg \) plank kept on a frictionless horizontal floor....

A man of mass 50kg50kg is standing on a 100kg100kg plank kept on a frictionless horizontal floor. Initially both are at rest. If the man starts walking on the plank with speed 6ms6\dfrac{m}{s} towards right relative to the plank, then amount of muscle energy spent by the man is

Explanation

Solution

Hint : To solve this question, we will be using the conservation laws for the given system and also we will have to consider the frame of reference of the man with respect to the plank as well as the frictionless surface.
(A) Conservation of angular momentum
M1V1=M2V2{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}
Where M1{{M}_{1}} and M2{{M}_{2}} are the masses. V1{{V}_{1}} and V2{{V}_{2}} are the velocities .
(B) Formula for kinetic energy
K=12MV2K=\dfrac{1}{2}M{{V}^{2}}
Where MM is the mass and VV is the velocity.

Complete Step By Step Answer:
Now for the first thing, we will be focusing on the data given.
Given, the mass of the man M1=50kg{{M}_{1}}=50kg
And the mass of the plank M2=100kg{{M}_{2}}=100kg
The frictional coefficient μ=0\mu =0
The velocity of the man with respect to the plank V1=6ms{{V}_{1}}=6\dfrac{m}{s}
The velocity of the plank with respect to the frictionless floor is V2{{V}_{2}}
Whereas, the velocity of the man with respect to the frictionless floor V=6ms−V2V=6\dfrac{m}{s}-{{V}_{2}}
Now, according to the data given to us, there is no external force acting on the system and therefore, the total momentum of the system is conserved. Hence, using the formulae for conservation of momentum
M1V=M2V2{{M}_{1}}V={{M}_{2}}{{V}_{2}}
⇒50(6−V2)=100V2\Rightarrow 50(6-{{V}_{2}})=100{{V}_{2}}
⇒300−50V2=100V2 ⇒150V2=300 ⇒V2=300150 \begin{aligned} & \Rightarrow 300-50{{V}_{2}}=100{{V}_{2}} \\\ & \Rightarrow 150{{V}_{2}}=300 \\\ & \Rightarrow {{V}_{2}}=\dfrac{300}{150} \\\ \end{aligned}
⇒V2=2ms\Rightarrow {{V}_{2}}=2\dfrac{m}{s}
Therefore, the velocity of the plank with respect to the frictionless surface is V2=2ms{{V}_{2}}=2\dfrac{m}{s} which is in the opposite direction to that of the velocity of the man. Because the floor is frictionless, when the man starts moving on the plank, the plank starts moving in the opposite direction.
Now, if we look at the figure closely, we will notice that the muscle energy which is spent by the man is equal to the total kinetic energy of the man and the plank system together.
We use the formula for kinetic energy as
K=12M1V2+12M2V2 ⇒K=12(50)(6−2)2+12(100)(2)2 ⇒K=600kg⋅m2s2×(Nkg⋅ms2) ⇒K=600J \begin{aligned} & K=\dfrac{1}{2}{{M}_{1}}{{V}^{2}}+\dfrac{1}{2}{{M}_{2}}{{V}^{2}} \\\ & \Rightarrow K=\dfrac{1}{2}(50){{(6-2)}^{2}}+\dfrac{1}{2}(100){{(2)}^{2}} \\\ & \Rightarrow K=600kg\cdot \dfrac{{{m}^{2}}}{{{s}^{2}}}\times (\dfrac{N}{kg\cdot \dfrac{m}{{{s}^{2}}}}) \\\ & \Rightarrow K=600J \\\ \end{aligned}
Finally, we can say that the energy used by the muscles of the man is equal to 600J600J .

Note :
Keep in mind to see the frame of reference while writing the relative velocities of man and plank with the frictionless floor. Also, remember that the velocity of the plan with respect to the man is in the opposite direction to that of the velocity of the man.