Question

The position $x$ of a particle moving along $x$-axis at time ($t$) is given by the equation, $t = \sqrt x + 2$, where $x$ is in meters and $t$ in second. Find the work done by the force in the first four seconds.
A. Zero
B. $2J$
C. $4J$
D. $8J$

Answer 1

Hint: According to the work energy theorem the total work done by all the forces on a body is equal to the change in kinetic energy of the body. Using the above relation of the position, find the relation of velocity by differentiating it with respect to time and then find kinetic energy before and after 4 seconds.

Complete answer:
We have a particle moving along $x$-axis with the following relation with respect to time, $t$:
$ \Rightarrow t = \sqrt x + 2$
$ \Rightarrow x = {(t - 2)^2} = {t^2} - 4t + 4$
Differentiating above equation with respect to $t$, we get equation for velocity along $x$-axis as follows:
$v = \dfrac{{dx}}{{dt}} = 2t - 4$ (in m/s).
Now we have an evolution equation for velocity along $x$-axis. Let the mass of the particle be $m$(in kg). Then kinetic energy of the particle moving along $x$-axis will be given as:
$K = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{(2t - 4)^2} = 2m{(t - 2)^2}$
According to the work energy theorem, the total amount of work done on an object because of all the forces is equal to the change in the kinetic energy of the body. Above problem asks us to calculate the work done on the body in the first 4 seconds of the motion. So we will find the difference in kinetic energy of the body at start and end of 4 seconds of the motion. At t=0, let the kinetic energy be $K1 = 2m{(0 - 2)^2} = 2m \times 4 = 8m$
And let the kinetic energy at the end of 4 seconds be $K2 = 2m{(4 - 2)^2} = 2m \times 4 = 8m$.
Work done by the forces on the body $ = K2 - K1 = 8m - 8m = 0$. So no work is done by the forces by moving the particle along the axis in the first 4 seconds.

Note: You can also solve the problem by substituting the values of position before and after four seconds. You will see that the value of $x$ is the same at both times. This means that the particle has returned back to the same point in four seconds. Since the total displacement is zero, we can say that total work done is zero.