Question

The masses ${M_1}$ and ${M_2}({M_2} > {M_1})$ are released from rest. Using work energy theorem, find out the velocity of the blocks when they move a distance x.

(A) $V = \sqrt {\dfrac{{({M_2} - {M_1})gx}}{{{M_1} + {M_2}}}}$
(B) $V = \sqrt {\dfrac{{2({M_2} - {M_1})gx}}{{{M_1} + {M_2}}}}$
(C) $V = \sqrt {\dfrac{{3({M_2} - {M_1})gx}}{{{M_1} + {M_2}}}}$
(D) $V = \sqrt {\dfrac{{4({M_2} - {M_1})gx}}{{{M_1} + {M_2}}}}$

Answer 1

Hint We should know that the work-energy theorem gives us an idea that the net work done by the forces on an object which is equal to the change in the kinetic energy of the body. Based on this theorem we have to answer this question.

Complete step by step answer
We should know that the expression for the diagram is given as:
${({W_{all}})_{system}} = {(\Delta K)_{system}}$
Now the expression is evaluated as:
${({W_g})_{sys}} + {({W_T})_{sys}} = {(\Delta K)_{sys}}$
In this case,
${({W_T})_{sys}} = 0$
So, the expression now is given as:
${m_2}gx - {M_1}gx = \dfrac{1}{2}({M_1} + {M_2}){V^2} - 0......(i)$
So, the value of V after the evaluating is given as:
$V = \sqrt {\dfrac{{2({M_2} - {M_1})gx}}{{{M_1} + {M_2}}}}$

So, the correct answer is Option B.

Note From the work-energy theorem we get an idea that the new work that is occurring on the object will cause a change in the kinetic energy of the object. The main formula is given as net work equals the change in the kinetic energy which is also similar to the difference between final kinetic energy and the initial kinetic energy.