Question

A body of mass 3 kg is under a force, which
GO
causes a displacement in it is given by S=t*3/3
(in metres). find the work done by the force
in first 2 seconds.​

Answer 1

SOLUTION.

As we know that work done is...

W = f.s,

where s is the displacement of the mass

f is the force acting on it.

But this formula is applicable only and only when force is constant as force varies. so this EQUATION turns into it.....

dW = F.dr

Where dr is the small displacement.

ACC to question

S = t³/3

differentiate both sides....

ds = t².dt....put in work equation.

again differentiate it

d²s = 2t.dt²

d²s/dt² = 2t

d²s/dt² = Acceleration

dW = F.t².dt........ F = ma

dW = m(2t)(t².dt).......... a = 2t

dW = 2mt³.dt

Integrate it.

∫ dW = 2m∫ t³.dt

Limit of work goes from 0 to w

Limit of time goes from 0 to 2

[ w - 0] = 2m[2⁴/4 - 0⁴/4]

w = 2m(4).......mass = 3kg

w = 24joules

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