
A body of mass starts moving from rest along x-axis so that its velocity varies as v=a$\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 $\dfrac{1}{8}m{a^4}{t^2}$
B $4m{a^4}{t^2}$
C $8m{a^4}{t^2}$
D $\dfrac{1}{4}m{a^4}{t^2}$
Answer
580.2k+ views
Hint: 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=$\dfrac{{dv}}{{dt}}$). Rate of change of displacement is called velocity (v=$\dfrac{{ds}}{{dt}}$). Displacement of a body moving with uniform acceleration is given as, S=ut+$\dfrac{1}{2}a{t^2}$.
Complete step by step answer:
Given, velocity v= a$\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’= $\dfrac{{dv}}{{dt}}$
Substituting the value of velocity v in the above equation,
a’=$\dfrac{{d\left( {a\sqrt s } \right)}}{{dt}}$
=$\dfrac{a}{{2\sqrt s }}\dfrac{{ds}}{{dt}}$
We know that $\dfrac{{ds}}{{dt}}$=v = a$\sqrt s $.
We get, a’=$\dfrac{a}{{2\sqrt s }} \times a\sqrt s $=$\dfrac{{{a^2}}}{2}$
Using the equation of uniform acceleration, we get displacement S=ut+$\dfrac{1}{2}a'{t^2}$.
Given that the initial velocity u=0 and a’=$\dfrac{{{a^2}}}{2}$$\dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}$
S=0×t+$\dfrac{1}{2} \times \dfrac{{{a^2}}}{2}{t^2} = \dfrac{1}{4}{a^2}{t^2}$
Therefore, work done=force×displacement
=ma’×S
=m$ \times \dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}$
=$\dfrac{1}{8}$ma$^4$t$^2$
So, the correct answer is “Option A”.
Note:
Students should learn a simple formula of differentiation and integration to solve such types of questions. Here Newton's law and equation for uniform acceleration are being used.
Additional information: Work is done when a force that is applied to an object moves that object. The work is calculated by multiplying the force by the amount of movement of an object (W = F×S).
Complete step by step answer:
Given, velocity v= a$\sqrt s $
Since work done is, W= force×displacement
According to Newton’s 2nd law of motion force
F= mass m× acceleration a’
Acceleration a’= $\dfrac{{dv}}{{dt}}$
Substituting the value of velocity v in the above equation,
a’=$\dfrac{{d\left( {a\sqrt s } \right)}}{{dt}}$
=$\dfrac{a}{{2\sqrt s }}\dfrac{{ds}}{{dt}}$
We know that $\dfrac{{ds}}{{dt}}$=v = a$\sqrt s $.
We get, a’=$\dfrac{a}{{2\sqrt s }} \times a\sqrt s $=$\dfrac{{{a^2}}}{2}$
Using the equation of uniform acceleration, we get displacement S=ut+$\dfrac{1}{2}a'{t^2}$.
Given that the initial velocity u=0 and a’=$\dfrac{{{a^2}}}{2}$$\dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}$
S=0×t+$\dfrac{1}{2} \times \dfrac{{{a^2}}}{2}{t^2} = \dfrac{1}{4}{a^2}{t^2}$
Therefore, work done=force×displacement
=ma’×S
=m$ \times \dfrac{{{a^2}}}{2} \times \dfrac{1}{4}{a^2}{t^2}$
=$\dfrac{1}{8}$ma$^4$t$^2$
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).
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