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Question: In the figure, \({{m}_{1}}=5kg\), \({{m}_{2}}=2kg\) and \(F=1N\). Find the acceleration of either bl...

In the figure, m1=5kg{{m}_{1}}=5kg, m2=2kg{{m}_{2}}=2kg and F=1NF=1N. Find the acceleration of either block. Describe the motion of m1m_1 if the string breaks but F continues to act.

Explanation

Solution

When an object is attached to a point with a string a tension force is connected to the string in the upward direction. When an object is hanging on a massless string a gravitational pull also acts on the body. The study of these forces can be done with the help of FBD (free body diagram) and their forces are used to balance the other forces to maintain the inertia of the object (mass).
As per the given data,
The mass of one block is m1=5kg{{m}_{1}}=5kg
The mass of another block is m2=2kg{{m}_{2}}=2kg
The force acting downward other than the gravitational pull is F=1NF=1N

Complete answer:
When we draw a free body diagram for both the blocks attached to the string.
The FBD of the block of mass m1{{m}_{1}} can be given as in the diagram here,

The force balancing equation of mass m1{{m}_{1}} can be written as,
T=m1gm1a+F m1a=m1g+FT...(1) \begin{aligned} & T={{m}_{1}}g-{{m}_{1}}a+F \\\ & \Rightarrow {{m}_{1}}a={{m}_{1}}g+F-T\quad ...\left( 1 \right) \\\ \end{aligned}
The FBD of the block of mass m2{{m}_{2}} can be given as in the diagram here,

The force balancing equation of mass m2{{m}_{2}} can be written as,
Tm2a=mg+F m2a=Tm2gF....(2) \begin{aligned} & T-{{m}_{2}}a=mg+F \\\ & \Rightarrow {{m}_{2}}a=T-{{m}_{2}}g-F\quad ....\left( 2 \right) \\\ \end{aligned}
Adding equation (1) and (2):
a=m1m2m1+m2a=\dfrac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}
By putting the values of m1{{m}_{1}} and m2{{m}_{2}}
a=10(37) a=4.2ms2 \begin{aligned} & a=10\left( \dfrac{3}{7} \right) \\\ & \therefore a=4.2m{{s}^{-2}} \\\ \end{aligned}
Thus, the acceleration acting on the blocks will be 4.2ms24.2m{{s}^{-2}}
When the string breaks the block will move towards the direction of applied force FF (Towards the ground). Then the forces acting on the block will be equated as,
m1a=F+m1g{{m}_{1}}a=F+{{m}_{1}}g
By putting the value of m1{{m}_{1}} the acceleration acting on the block will be given as,
5a=1+5g a=g+0.2ms2 \begin{aligned} & \Rightarrow 5a=1+5g \\\ & \therefore a=g+0.2m{{s}^{-2}} \\\ \end{aligned}
Thus, the acceleration of mass m2{{m}_{2}}and m1{{m}_{1}} will be the same when they are attached to the string and the value of acceleration is 4.2ms24.2m{{s}^{-2}}. When the string breaks the acceleration of the block m1{{m}_{1}} will be g+0.2ms2g+0.2m{{s}^{-2}}. And these are the required answers to the question.

Note:
When an object attached to the string is facing a certain force and breaks after a certain period of time its acceleration will increase to balance the force. This increase in energy is due to the conversion of the potential energy that the object holds at a certain height to the kinetic energy. This conversion gives proof for the law of conservation of energy.