Question

A body of mass \[3Kg\]is under a force, which causes a displacement in it is given by \[S=\dfrac{{{t}^{3}}}{3}\] (in \[m\]). Find the work done by the force in the first\[2\sec \].

Answer 1

Hint: When attempting this question, keep in mind that \[Work\] is done when \[force\] is applied to an object the object moves. \[Work\] is calculated by multiplying the amount of movement (here it will be the displacement) to the amount of \[force\] exerted on the object to make it move.

Complete step-by-step answer:
Work is an important quantity in physics. It is defined as the energy transferred from one type to another whenever a force is applied which causes the object to move a certain distance. In equation form, work is a force multiplied by the displacement moved in the direction of that of force.
If force is constant then finding work is not difficult, but if the force is constantly changing as what we have been given in our question, finding the work done becomes a little bit more complex. And here is when you need a more fundamental equation for work. Work isn’t just a force multiplied by displacement here, it is the integral of the force over the displacement. In other words if you have force as a function of displacement and integrate it keeping in mind displacement, you will get the expression of work.
Which when we derive, the expression will come out to be
 \[w=\int{fd}\]
We are given the values of the mass \[m\]which is \[3Kg\]and we are given the relationship between the displacement it undergoes due to the force and the time it takes to do that. Now we need to find out the work done by that force on the mass in the first \[2\]seconds.
From question we are given;
\[S=\dfrac{{{t}^{3}}}{3}\]
Where \[S\]is the displacement and \[t\] is the time.
Differentiating the equation given we get;
\[\Rightarrow \dfrac{ds}{dt}=v={{t}^{2}}\]
\[\Rightarrow a=2t\]
We know that, \[w=\int{fd}\]
\[=\int\limits_{0}^{2}{ma{{t}^{2}}dt}\]
\[\Rightarrow 3\int\limits_{0}^{2}{2{{t}^{3}}dt}\]
\[\Rightarrow 6\int\limits_{0}^{2}{{{t}^{3}}dt}\]
\[\Rightarrow \dfrac{6}{4}\left[ {{t}^{4}} \right]_{0}^{2}\]
\[\Rightarrow \dfrac{6}{4}\times 16=24J\]
\[\therefore Work=24J\]
Hence the work done by the force on the object to make it move in the first \[2\]seconds is \[24J\].

Note: Energy is defined as the capacity to do work. It is the strength to do any kind of physical activity. Resources are processed to get the energy that is used to provide light or heat. Energy can never be created nor destroyed, and can only be transformed from one form to another.