Question

A body is acted by a force such that the displacement is related with time \[t\] as \[t = \sqrt x + 3\]. Find work-done by this force in \[0\] to \[6\sec \].

Answer 1

Hint:Relation between displacement and time is given, we can find the value of velocity by using it at time =0 and 6 sec. After finding the value of velocity, we can substitute it in the work-done formula.

Complete step by step answer:
Displacement is related with time, given-\[t = \sqrt x + 3\]
Here t is time and x is the displacement of the body on which force is acting.
For finding the work-done, we must calculate the value of initial and final velocity. So, we can modify the given equation as-
\[\sqrt x = t - 3\]
Take square on both sides of this equation, we get-
\[x = {\left( {t - 3} \right)^2}\]
\[\Rightarrow x = {t^2} + 9 - 6t\] ----- (1)
Now this x is the displacement equation. For finding velocity, we know that velocity is the rate of change of displacement. So,
\[v = \dfrac{{dx}}{{dt}}\]
Now, differentiate equation (1), we get-
\[v = \dfrac{{d\left( {{t^2} + 9 - 6t} \right)}}{{dt}}\]
\[\Rightarrow v = 2t - 6\] --- (2)
Now, substitute the value of time in equation (2),
At t=0, \[v = 6m/s\]
At \[t = 6\sec \],\[6m/s\]
Work-done= change in kinetic energy=\[K.{E_f} - K.{E_i}\]
Here \[K.{E_f} = \]final kinetic energy
\[K.{E_i} = \]Initial kinetic energy
Work-done=\[\dfrac{1}{2}m{\left( 6 \right)^2} - \dfrac{1}{2}m{\left( 6 \right)^2}\]
$\therefore$ Work-done=0

Hence, work-done is zero.

Note:The kinetic energy of any object depends on its velocity. If we want to change velocity, then some force must act on it.
W=\[K.{E_f} - K.{E_i}\]
While substituting the value of initial and final kinetic energies, remember to take the velocities for the initial and final time. In this question, both velocities are the same, that's why work done is zero.