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Question: The position of a particle as a function of time is described by relation \(x = 3t - 3t^2 + t^3\) wh...

The position of a particle as a function of time is described by relation x=3t−3t2+t3x = 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

Explanation

Solution

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=∫FdxW = \int F dx

Complete step by step answer:
First we write down the expression for position x:
x=3t−3t2+t3x = 3t - 3t^2 + t^3 .
By differentiating this expression twice w.r.t. time we get acceleration as:
a=d2xdt2=−6+6ta = \dfrac{d^2 x}{dt^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+60tF = -60 + 60t
Infinitesimal work done due to this force causing displacement dx can be written as:
dW=F.dxdW = F.dx
Therefore, we need to determine dx first. From the expression for x, we can write:
dx=3dt−6tdt+3t2dtdx = 3 dt - 6t dt + 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+3t2dt)F.dx = (-60 + 60t).(3 dt - 6t dt + 3t^2 dt)
F.dx=−180dt+360tdt−180t2dt+180tdt−360t2dt+180t3dtF.dx = -180 dt + 360t dt -180t^2 dt + 180t dt -360t^2 dt + 180t^3 dt
F.dx=−180dt+540tdt−540t2dt+180t3dtF.dx = -180 dt + 540t dt -540t^2 dt + 180t^3 dt
Performing integration now, we get the work done as:
W=∫03(−180+540t−540t2+180t3)dtW = \int_{0}^{3} (-180 + 540t -540t^2 + 180t^3 )dt
we kept the limits t = 0 s to t = 3 s.
W=−180(t)03+540(t22)03−540(t33)03+180(t44)03W = -180 (t)_{0}^{3}+ 540 \left( \dfrac{t^2}{2} \right)_{0}^{3} -540 \left( \dfrac{t^3}{3} \right)_{0}^{3} + 180 \left( \dfrac{t^4}{4} \right)_{0}^{3}
W=−540+2430−4860+3645=675JW = -540 + 2430 - 4860 + 3645 = 675 J

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.