Question

Under the action of a force, a 1 kg body moves, such that its position x as a function of time t is given by x =t3/2, where x is in meter and t is in second. the work done by the force in first 3 second is

Answer 1

Differentiating in this question would be lengthy so we can use the formula
work done = change in kinetic energy.
But we have to calculate velocity for this,
and given is the relation between x&t
$x = {t}^{ \frac{3}{2} }$
differentiating
$\frac{dx}{dt} = \frac{3}{2} {t}^{ \frac{1}{2} }$
i.e. velocity.
Now
$ki = \frac{1}{2} m {vi^{2}}$
at t=0
vi=0
$kf = \frac{1}{2} m {vf}^{2}$
$vf = \frac{3}{2} \times {3}^{ \frac{1}{2} }$
now inserting value of vf in kf we get
$kf = \frac{27}{8}$

work done=27/8 J

Answer 2

Explanation:

Given → mass = 1 kg

X = (t^3)/2

differentiate w.r.t. time

dX/dt = 3t²/2

dX = 3t²/2 * dt -------1)

V = 3t²/2

differentiate again w.r.t. time

A = 6t/2 = 3t

F = mA

F = 1 kg * 3t = 3t N

W = integration of [ F.dX]

= integration [3t * 3t²/2 dt ] -----from 1)

= (9/2) integration of [t^3 dt]

= 9/2 [t^4 / 4 ]

$w \: = \frac{9}{4} t {}^{4}$

put , t = 3

$w \: = \frac{729}{4}$ J