Question

A time-dependent force $F = 6t$ acts on a particle of mass$1kg$ . If the particle starts from rest, the work done by the force during the first $1\sec $ will be:
$\left( a \right)$ $18J$
$\left( b \right)$ $4.5J$
$\left( c \right)$ $22J$
$\left( d \right)$ $9J$

Answer 1

Hint Since the force is directly proportional to time so by increasing the time force will also increase. Firstly the particle is at rest but now it will have some particular velocity as there is the force acting on it. So by using the formula of the force and differentiating and integrating, we would be able to get the work done by the force acting for that particular time.
Formula used
Force,
$F = ma$
Here,
$F$, will be the force acting on it
$m$, will be the mass of the body
$a$, will be the acceleration
Also from the work-energy theorem,
$W = \vartriangle KE = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)$
Here,
$W$, will be the work done
$\vartriangle KE$, will be the change in the kinetic energy
$v$, will be the final velocity
$u$, will be the initial velocity

Complete Step By Step Solution As we know from the formula,
$F = ma$
And we can write acceleration in the form of a derivative.
$ \Rightarrow m \times \dfrac{{dv}}{{dt}}$
So from the question,
$ \Rightarrow m\dfrac{{dv}}{{dt}} = 6t$
Taking $dt$to the right side, we get
$ \Rightarrow m \times dv = 6t \times dt$
Solving for the$dv$, we get
$ \Rightarrow dv = 6t \times dt$
Now, we will integrate the above equation
$ \Rightarrow \int\limits_0^v {dv} = \int\limits_0^1 {6tdt} $
By applying the integration formula,
$ \Rightarrow v = 6\left[ {\dfrac{{{t^2}}}{2}} \right]_0^1$
After solving the equation, we will get the following
$ \Rightarrow 6 \times \dfrac{1}{2}$
And again solving it, we get
$ \Rightarrow 3m/s$
Now from the work-energy theorem,
We know, work done will be equal to a change in the kinetic energy,
Mathematically, it can be written as
$W = \vartriangle KE = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)$
So, we will substitute the values we had got above
$ \Rightarrow \dfrac{1}{2} \times 1\left( {{3^2} - {0^2}} \right)$
And on solving the above equation, we get
$ \Rightarrow 4.5J$

Therefore, $4.5J$ work is done$1\sec $.

Note Work-Energy Theorem cares with the amendment of K.E. of an object because of work done by the net force functioning on an object. The net force is the resultant of all forces functioning on the object, and would usually embody gravity. Conservation of energy cares with the amendment of energy in a system because of work done by external forces.