Question

A force acts on a 30g particle in such a way that the position of the particle as a function of time is given, $x = 3t - 4{t^2} + {t^3}$. where $x$ is in meters and $t$ is in second. The work done during the first 4 seconds is:
A. $5.28J$
B. $450mJ$
C. $490mJ$
D. $530mJ$

Answer 1

Hint: Whenever a force is applied on a body, it causes an effect in the body. The quantity of work done is the product of work done and the displacement caused due to the effect of the force.
Work done, W = $F \times S$
where F = force, S = displacement.
The quantity work is a scalar and the units of the work in SI units is joules (J).

Complete step-by-step answer:
When a force F is applied on the body and it moves by the distance S, the work is said to be done on the body and this work is equal to:
$W = F \times S$
The force applied on the body,
$F = ma$
Given, mass of the object, m = 30 g
Displacement function, $x\left( t \right) = 3t - 4{t^2} + {t^3}$
Acceleration is defined as the rate of change of velocity with time, which in turn, is nothing but the rate of change of speed in time.
If we were to apply that in respect to the functions, we would obtain the relation that the acceleration can be calculated by differentiating the displacement function twice.
$
  a = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dx}}{{dt}}} \right) \\
  x = 3t - 4{t^2} + {t^3} \\
  \dfrac{{dx}}{{dt}} = 3 - 8t + 3{t^2} \\
  \dfrac{{{d^2}x}}{{d{t^2}}} = - 8 + 6t \\
 $
Mass, m = $30g = 0.03kg$

Here, the work done is equal to:
$
  dW = F.dx \\
   \to dW = ma.x(t).dt \\
 $
Integrating the above in the first 4 seconds will give us the work done in 4 seconds.
Work done in the first four seconds,
\[
  W = \int\limits_0^4 {Fx(t).dt} \\
   \to W = \int\limits_0^4 {\left( {ma} \right)x(t).dt} \\
   \to W = 0.03\int\limits_0^4 {\left( { - 8 + 6t} \right)\left( {3t - 4{t^2} + {t^3}} \right).dt} \\
   \to W = 0.03\int\limits_0^4 {18{t^3} - 72{t^2} + 82t - 24.dt} \\
   \to W = 0.03\left( {\dfrac{{18{t^4}}}{4} - \dfrac{{72{t^3}}}{3} + \dfrac{{82{t^2}}}{2} - 24t} \right)_0^4 \\
 \]
Solving the above definite integral, we get –
\[
  W = 0.03\left( {\dfrac{{18 \times {4^4}}}{4} - \dfrac{{72 \times {4^3}}}{3} + \dfrac{{82 \times {4^2}}}{2} - 24 \times 4 - 0} \right) \\
   \to W = 0.03\left( {1152 - 1536 + 656 - 96} \right) \\
   \to W = 0.03\left( {176} \right) \\
  \therefore W = 5.28J \\
 \]

Hence, the correct option is Option A.

Note: In this problem, the differentiation and integration of the variables have been performed based on some rules. Refer to these rules always to understand the process of differentiation and integration better.