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Question: In the diagram shown, no friction at any contact surface. Initially the spring has no deformation, w...

In the diagram shown, no friction at any contact surface. Initially the spring has no deformation, what will be the maximum deformation in the spring considering all the strings to be sufficiently large consider the spring constant to be K.

A) 8F/3K8F/3K
B) 9F/3K9F/3K
C) 105F/3K105F/3K
D) 98F/3K98F/3K

Explanation

Solution

Using the theory of the system reaching a state of rest at maximum separation condition on the work done in the center of mass frame, we will deduce a formula with displacement. Then using the relation of center of mass of the two bodies in a system, we find out the displacement of the body 11. Upon substitution of the determined data, we will find the value of maximum displacement.

Formulae used:
Work done in the center of mass frame in terms of weight of the bodies: Wspring+WF+W2F=0{W_{spring}} + {W_F} + {W_{2F}} = 0.
Where Wspring,WF,W2F{W_{spring}},{W_F},{W_{2F}} are the work done of the spring, 2M2M block and MM block and is expressed in Joules (J)(J).
Work done in the center of mass frame in terms of kinetic energy, displacement and force of the bodies: 12kx2max+Fx1+2Fx2=0 - \dfrac{1}{2}k{x^2}_{max} + {F_{x1}} + 2{F_{x2}} = 0
Where kk is the spring constant, x1,x2{x_1},{x_2} are the displacements of the 2M2M block and MM block and is expressed in meter (m)(m) and FF is the force acting on the system and is expressed in Newton (N)(N).
Relation of center of mass of the two bodies in a system: Mx1=2Mx2{M_{x1}} = 2{M_{x2}}
Where Mx1,Mx2{M_{x1}},{M_{x2}} are the masses of the 2M2M and MM block and are expressed in kilograms (kg)(kg).

Complete step by step solution:


In the given system, the force acting on the block of mass 2M2M will have a force of 2F2F acting on it while the block MM will have force FF acting on it. On consideration of conditions in a state of maximum separation of the two blocks, the entire system comes to rest for a particular instant of time.

During this, the work done in the center of the mass frame will be zero. Therefore we get the total work done equation as Wspring+WF+W2F=0{W_{spring}} + {W_F} + {W_{2F}} = 0. But, work done is a product of force and displacement for bodies and kinetic energy of springs.

Therefore,
Wspring+WF+W2F=0 12kxmax2+Fx1+2Fx2=0  {W_{spring}} + {W_F} + {W_{2F}} = 0 \\\ \Rightarrow \dfrac{1}{2}kx_{max}^2 + F{x_1} + 2F{x_2} = 0 \\\
Cancelling out the force factors we get,
xmax=x1+x2{x_{max}} = {x_1} + {x_2}
According to the given data, the masses of the two bodies are MM and 2M2M.
Therefore we can write it as Mx1=2Mx2{M_{x1}} = 2{M_{x2}}
Upon reduction we get,

Mx1=2Mx2 x1=2x2  {M_{x1}} = 2{M_{x2}} \\\ \Rightarrow {x_1} = 2{x_2} \\\

Substituting the value of x1{x_1} in the displacement equation we get,

2x2+x2=xmax x2=xmax3,x1=23xmax  2{x_2} + {x_2} = {x_{max}} \\\ {x_2} = \dfrac{{{x_{max}}}}{3},{x_1} = \dfrac{2}{3}{x_{max}} \\\

Substituting these values of x1{x_1} and x2{x_2} in the total energy equation of the system at rest we get,
12kxmax2+Fx1+2Fx2=0 12kxmax2+F(23xmax)+2F(xmax3)=0 12kxmax2=4F3xmax xmax=8F3k  \dfrac{1}{2}kx_{max}^2 + F{x_1} + 2F{x_2} = 0 \\\ \Rightarrow \dfrac{1}{2}kx_{max}^2 + F(\dfrac{2}{3}{x_{max}}) + 2F(\dfrac{{{x_{max}}}}{3}) = 0 \\\ \Rightarrow \dfrac{1}{2}kx_{max}^2 = \dfrac{{4F}}{3}{x_{max}} \\\ \Rightarrow {x_{max}} = \dfrac{{8F}}{{3k}} \\\
In conclusion, the correct option is A.

Note: We must consider the boundary condition of rest at maximum separation. Because at that point, the whole system becomes static and then due to restoring force of the spring, it starts regaining its original position. Also, the work done by spring is negative as it is opposite to the direction of the separation pull force.