Question:

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36\, kg$ and $0.72\, kg$. Taking $g =10\, m / s ^{2}$, find the work done (in joules) by the string on the block of mass $0.36\, kg$ during the first second after the system is released from rest.

Updated On: Jan 30, 2025
  • 4 J
  • 2 J
  • 8 J
  • 10 J
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The Correct Option is C

Solution and Explanation

Given $m =0.36\, kg,\, M =0.72\, kg.$
The figure shows the forces on $m$ and $M$.
When the system is released, let the acceleration be $a$. Then
Given $m =0.36\, kg,\, M =0.72\, kg$.
The figure shows the forces on $m$ and $M$.
When the system is released, let the acceleration be $a$. Then
$T=m g=m a$
$M g-T=M$
$\therefore a=\frac{(M-m) g}{M +m}=b/3$ and $T=4\, m g/3 $ (For block $m$)
$u=0, a=g/3, t=1, s=?$
$s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times \frac{g}{3} \times 1^{2}=g/6$
$\therefore$ Work done by the string on $m$ is
$\vec{T} \vec{s}=T s=4 \frac{m g}{3} \times \frac{g}{6}$
$=\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}$
$=8 J$
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Concepts Used:

Newtons Laws of Motion

Newton’s First Law of Motion:

Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.

Newton’s Second Law of Motion:

Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.

Mathematically, we express the second law of motion as follows:

Newton’s Third Law of Motion:

Newton’s 3rd law states that there is an equal and opposite reaction for every action.