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Question: A body of mass starts moving from rest along x-axis so that its velocity varies as v=a\(\sqrt s \)wh...

A body of mass starts moving from rest along x-axis so that its velocity varies as v=as\sqrt s where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first seconds after the start of the motion is:
A 18ma4t2\dfrac{1}{8}m{a^4}{t^2}
B 4ma4t24m{a^4}{t^2}
C 8ma4t28m{a^4}{t^2}
D 14ma4t2\dfrac{1}{4}m{a^4}{t^2}

Explanation

Solution

The work is calculated by multiplying the force by the amount of movement of an object (W = F × d). According to Newton's second law of motion force is defined as the product of mass and acceleration (F=m×a). Acceleration can be defined as rate of change of velocity (a=dvdt\dfrac{{dv}}{{dt}}). Rate of change of displacement is called velocity (v=dsdt\dfrac{{ds}}{{dt}}). Displacement of a body moving with uniform acceleration is given as, S=ut+12at2\dfrac{1}{2}a{t^2}.

Complete step by step answer:
Given, velocity v= as\sqrt s
Since work done is, W= force×displacement
According to Newton’s 2nd law of motion force
F= mass m× acceleration a’
Acceleration a’= dvdt\dfrac{{dv}}{{dt}}
Substituting the value of velocity v in the above equation,
a’=d(as)dt\dfrac{{d\left( {a\sqrt s } \right)}}{{dt}}
=a2sdsdt\dfrac{a}{{2\sqrt s }}\dfrac{{ds}}{{dt}}
We know that dsdt\dfrac{{ds}}{{dt}}=v = as\sqrt s .
We get, a’=a2s×as\dfrac{a}{{2\sqrt s }} \times a\sqrt s =a22\dfrac{{{a^2}}}{2}
Using the equation of uniform acceleration, we get displacement S=ut+12at2\dfrac{1}{2}a'{t^2}.
Given that the initial velocity u=0 and a’=a22\dfrac{{{a^2}}}{2} a22×14a2t2\dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}
S=0×t+12×a22t2=14a2t2\dfrac{1}{2} \times \dfrac{{{a^2}}}{2}{t^2} = \dfrac{1}{4}{a^2}{t^2}
Therefore, work done=force×displacement
=ma’×S
=m×a22×14a2t2 \times \dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}
=18\dfrac{1}{8}ma4^4t2^2

So, the correct answer is “Option A”.

Note:
Students should learn a simple formula of differentiation and integration to solve such types of questions. Here Newton's law and equation for uniform acceleration are being used.
Additional information: Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F×S).