Question

The position of a particle as a function of time is described by relation $x=3t-3t^2+t^3$ where the quantities are expressed in S.I. units. If mass of the particle be 10 kg, the work done in first three seconds is
A. 10 J
B. 30 J
C. 300 J
D. 675 J

Answer 1

Hint: We are given position x in terms of another variable t (time), we must differentiate this twice to get the acceleration equation which when multiplied by mass will give us a force equation. Then we can use Force times displacement to get the work done.

Formula used:
For a small displacement dx, caused by a force F the work done is given as:
$W=\int Fdx$

Complete step by step answer:
First we write down the expression for position x:
$x=3t-3t^2+t^3$ .
By differentiating this expression twice w.r.t. time we get acceleration as:
$a={\displaystyle \frac{d^{2}x}{d{t}^2}}=-6+6t$.
We can now write the expression for force F = ma; as we know mass of the particle is 10 kg so;
$F=-60+60t$
Infinitesimal work done due to this force causing displacement dx can be written as:
$dW=F.dx$
Therefore, we need to determine dx first. From the expression for x, we can write:
$dx=3dt-6tdt+3t^{2}dt$
where we simply differentiate once on both sides of x with respect to time.
Multiply the dx with F first to simplify the integration ahead:
$F.dx=(-60+60t).(3dt-6tdt+3t^{2}dt)$
$F.dx=-180dt+360tdt-180t^{2}dt+180tdt-360t^{2}dt+180t^{3}dt$
$F.dx=-180dt+540tdt-540t^{2}dt+180t^{3}dt$
Performing integration now, we get the work done as:
$W={\int }_0^3(-180+540t-540t^2+180t^3)dt$
we kept the limits t = 0 s to t = 3 s.
$W=-180(t{)}_0^3+540{\left({\displaystyle \frac{t^2}{2}}\right)}_0^3-540{\left({\displaystyle \frac{t^3}{3}}\right)}_0^3+180{\left({\displaystyle \frac{t^4}{4}}\right)}_0^3$
$W=-540+2430-4860+3645=675J$

Therefore, the correct answer is option (D). 675 J is the work done in the first 3 seconds.

Note:
The integration in the formula for work usually is done considering the variable to be x (as we had F.dx) but here we made a variable change from x to t with the help of the given expression for x. As we change the variable from x to t we also change the limits accordingly. As we had to find work done for the first 3s, we took initial time t = 0s and final time to be t = 3s in the limits of integral.