Question

A force acts on a $3g$ particle in such a way that the position of the particle as a function of time is given, ${\text{x = 3t - 4}}{{\text{t}}^2}{\text{ + }}{{\text{t}}^3}$. where $x$ is in meters and t is in second. The work done during the first 4 seconds is:
A. $2.88{\text{ J}}$
B. ${\text{450 mJ}}$
C. ${\text{490 mJ}}$
D. ${\text{530 mJ}}$

Answer 1

Hint:Work is said to be done if force acting on a body displaces the body through a certain distance and the force has some component along the displacement. If a force $F$ acting on a body produces a displacement s in the body, then the work done $W$ is given by $W{\text{ = }}F\cos \theta {\text{ }}{\text{. }}s{\text{ = }}Fs\cos \theta {\text{ = }}\vec F{\text{ }}{\text{. }}\vec s$ where $\theta $ is the angle which the force makes with the direction of the displacement.

Complete step by step answer:
It is given mass $ = {\text{ 3 g = 0}}{\text{.003 kg}}$
Displacement ${\text{ = x}}\left( {\text{t}} \right){\text{ = 3t - 4}}{{\text{t}}^2}{\text{ + }}{{\text{t}}^3}$
We know work done by a constant force is given by,
$W{\text{ = }}F{\text{ }}{\text{. }}s$
Where we can write force as the product of mass and acceleration that is,
${\text{F = ma}}$
Mass is given and we want acceleration to get acceleration we have given a displacement function. So, by differentiating the displacement function twice we will get the acceleration.
$a{\text{ = }}\dfrac{{{d^2}x}}{{d{t^2}}}{\text{ = }}\dfrac{d}{{dx}}\left( {\dfrac{{dx}}{{dt}}} \right)$
${\text{x = 3t - 4}}{{\text{t}}^2}{\text{ + }}{{\text{t}}^3}$

Differentiating x with respect to dt
$\dfrac{{dx}}{{dt}}{\text{ = 3 - 8t + 3}}{{\text{t}}^2}$
Differentiating x again with respect to dt
$\dfrac{{{d^2}x}}{{d{t^2}}}{\text{ = - 8 + 6t}}$
So work done is equal to,
$dW{\text{ = }}F{\text{. }}dx$
Which can be written as,
$ \Rightarrow dW{\text{ = ma}}{\text{.x}}\left( {\text{t}} \right).dt$
By integrating the above equation in the first 4 seconds we will get the work done in the first 4 seconds.Therefore, work done in the first four seconds is equal to
$W = \int\limits_0^4 {Fx\left( {\text{t}} \right)} .dt$

By, rewriting the above terms we get
$ \Rightarrow W = \int\limits_0^4 {\left( {ma} \right)x\left( {\text{t}} \right)} .dt$
Taking constant outside
$ \Rightarrow W = m\int\limits_0^4 {\left( a \right)x\left( {\text{t}} \right)} .dt$
Substituting the values we get,
$ \Rightarrow W = 0.003\int\limits_0^4 {\left( { - 8 + 6t} \right)\left( {{\text{3t - 4}}{{\text{t}}^2}{\text{ + }}{{\text{t}}^3}} \right)} {\text{ }}dt$
$ \Rightarrow W = 0.003\int\limits_0^4 {\left( {18{t^3} - 72{t^2} + 82t - 24} \right)} {\text{ }}dt$
Upon simplifying we get
$ \Rightarrow W = 0.003\left( {\dfrac{{18{t^4}}}{4} - \dfrac{{72{t^3}}}{3} + \dfrac{{82{t^2}}}{2} - 24t} \right)_0^4$

On solving the definite integral, we get
$W = 0.003\left\{ {\left( {\dfrac{{18 \times {4^4}}}{4} - \dfrac{{72 \times {4^3}}}{3} + \dfrac{{82 \times {4^2}}}{2} - 24 \times 4} \right) - \left( 0 \right)} \right\}$
$ \Rightarrow W = 0.003\left( {1152 - 1536 + 656 - 96} \right)$
On simplifying we get,
$ \Rightarrow W = 0.003\left( {176} \right)$
$\therefore W = 0.528{\text{ J}}$
Which is nearly equal to $530{\text{ mJ}}$.

Therefore, the correct answer is option D.

Note:In calculating the work, a force does on an object as the object moves through some displacement, we use only the force component along the object's displacement. The force component perpendicular to the displacement does zero work. Whenever the component of force parallel to the displacement is oppositely directed to the displacement, the work done becomes negative. When an external agent causes a body to move in the direction of the force acting on the body. Such a work is termed as positive work. Work is a scalar quantity. The dimension of work is $\left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^2}} \right]$. In the SI system the absolute unit of work is joule. In the CGS system the absolute unit of work is ergs. $1{\text{ J = 1}}{{\text{0}}^7}{\text{ erg}}$