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Standard XII

Physics

Elastic Potential Energy

Two blocks of...

Question

Two blocks of masses $3$ kg and $2$ kg connected by a rope of mass $\frac{1}{2}$ kg are resting on a frictionless floor. If a constant force of $22$ N is applied to teh heavier block, then the tension in the mid of string will be.

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Solution

The correct option is A $9$ N
Total mass of the system $= 3 + 2 + \frac{1}{2} = \frac{11}{2}$ kg
Given, applied force $= 22$ N
So, the acceleration of the system is $a = \frac{F_{a p p}}{m} = \frac{22}{11 / 2} = 4 m / s^{2}$
At the mid of string, the applied force pulling the mass
$m^{'} = \frac{(\frac{1}{2})}{2} + 2 = \frac{9}{4}$ kg
Hence, tension in the middle of string will be
$T = m^{'} a = \frac{9}{4} \times 4 = 9$ N.

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Two blocks of masses 3kg and 2kg connected by a rope of mass 12kg are resting on a frictionless floor. If a constant force of 22N is applied to teh heavier block, then the tension in the mid of string will be.

The correct option is A 9NTotal mass of the system =3+2+12=112kgGiven, applied force=22NSo, the acceleration of the system is a=Fappm=2211/2=4m/s2At the mid of string, the applied force pulling the massm′=(12)2+2=94kgHence, tension in the middle of string will beT=m′a=94×4=9N.

Two blocks of masses $3$ kg and $2$ kg connected by a rope of mass $\frac{1}{2}$ kg are resting on a frictionless floor. If a constant force of $22$ N is applied to teh heavier block, then the tension in the mid of string will be.