Question

A locomotive of mass $m$ starts moving so that its velocity varies according to the law $v=a\sqrt{s}$, where $a$ is a constant, and $s$ is the distance covered. The total work performed by all the forces which are acting on the locomotive during the first $t$ seconds after the beginning of motion is $W={\displaystyle \frac{m{a}^4{t}^2}{x}}$. Find $x$.

Answer 1

Hint: The formula for the acceleration in terms of the velocity of the body must be applied and the use of the given equation for velocity will lead to the determination of the acceleration of the locomotive. The formula for the force that is exerted on the locomotive is applied and hence the work done is obtained from this force equation.

Complete step by step answer:
The above problem asks us to determine the value of the variable $x$ from the work done equation by a locomotive. This can be determined by finding out the acceleration of the locomotive and hence determining the relation between this acceleration and the work done by the locomotive.

We are given the equation for velocity of the vehicle known as a locomotive in terms of the distance that is covered by the locomotive. This equation is given to us as:
$v=a\sqrt{s}$ -----------($1$)
Locomotive is a vehicle which is similar to a train and hence consists of an engine as well and is said to be self-propelled which means the motion of the locomotive is automatic without any manual intervention and is used for dragging trains or wagons along with it.

Since, the locomotive is said to move on its own it will consist of a certain velocity with which it moves and thus an acceleration that is associated with it since the velocity is varying constantly. We are asked to find the value of $x$ from the equation for the work done by the locomotive and in order to do this we first need to derive this work done equation so that the derived equation can be equated with the given work equation to obtain the value of $x$.

As our first step we need to find the acceleration of the locomotive. Acceleration is said to be the rate of change of velocity with time and the change of the quantity considered mathematically by performing differentiation and hence the velocity is differentiated with respect to time. Thus we have:
$A={\displaystyle \frac{dv}{dt}}$
Where, $A$ indicates the acceleration of the locomotive.

We now substitute the equation for velocity from equation ($1$) in the above equation for acceleration of the locomotive and hence we get:
$A={\displaystyle \frac{d}{dt}}\left[a\sqrt{s}\right]$
Here, since the variable $a$ is said to be given as a constant it can be taken out as differentiating is only performed for variables and not constants. Hence, we get:
$A=a{\displaystyle \frac{d}{dt}}\left[\sqrt{s}\right]$
We can also write the above equation as:
$A=a{\displaystyle \frac{d}{dt}}\left[s^{{\displaystyle \frac{1}{2}}}\right]$

After performing differentiation as per its basic rules and also by applying the chain rule for differentiation we get:
$A={\displaystyle \frac{1}{2}}a\times s^{{\displaystyle \frac{1}{2}}-1}\times {\displaystyle \frac{d}{dt}}\left[s\right]$
On simplifying further we get:
$A={\displaystyle \frac{1}{2}}a\times s^{-{\displaystyle \frac{1}{2}}}\times {\displaystyle \frac{d}{dt}}\left[s\right]$
$\Rightarrow A={\displaystyle \frac{a}{2s^{{\displaystyle \frac{1}{2}}}}}{\displaystyle \frac{d}{dt}}\left[s\right]$
By rearranging the terms we get:
$A={\displaystyle \frac{a}{2\sqrt{s}}}{\displaystyle \frac{ds}{dt}}$ ---------($2$)

However we already know that the velocity is also defined as the rate of change of position or distance with time and hence the equation is given as:
$v={\displaystyle \frac{ds}{dt}}$
Also, we are given the equation for velocity in the question and thus from equation ($1$) we know that:
$v=a\sqrt{s}$
Hence, by equation these two equations for velocity we get:
${\displaystyle \frac{ds}{dt}}=a\sqrt{s}$
Hence, we substitute the above value in the equation ($2$) to get:
$A={\displaystyle \frac{a}{2\sqrt{s}}}\left(a\sqrt{s}\right)$
This equation is solved out and the common terms are cancelled out to get:
$A={\displaystyle \frac{a^2}{2}}$ ------------($3$)
Thus, this is the equation for the acceleration of the locomotive.

Now, if there is acceleration that is involved there will be a force that comes into picture due to which there is acceleration that is produced. This is in accordance to the Newton’s second law which gives the equation below:
$F=ma$
Here, the acceleration is represented by a variable $A$ and hence the equation becomes:
$F=mA$ --------------($4$)
The equation that we had determined for the acceleration of the locomotive is substituted in equation ($4$) and hence by substituting equation ($3$) in equation ($4$) we get:
$F=m\left[{\displaystyle \frac{a^2}{2}}\right]$
By rearranging the terms we have:
$F={\displaystyle \frac{m{a}^2}{2}}$ ------------($5$)

Since, the locomotive, similar to a train travels along a track in a straight line motion we can apply one of the equation for kinetics and hence we have:
$s=ut+{\displaystyle \frac{1}{2}}A{t}^2$
Here, the initial velocity of the locomotive will be zero because initially the locomotive will be at rest and when a body is at rest or when it is stationary then its velocity is said to be zero and thus $u=0$. Hence we get:
$s=\left(0\right)t+{\displaystyle \frac{1}{2}}a{t}^2$
$\Rightarrow s={\displaystyle \frac{1}{2}}A{t}^2$
Thus this is the equation for the distance covered by the locomotive. Therefore, by substituting the equation ($3$) in the above distance equation we get:
$s={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{a^2}{2}}\right]t^2$
Hence on solving it further we get:
$s={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{a^2{t}^2}{2}}$
$\Rightarrow s={\displaystyle \frac{a^2{t}^2}{4}}$ -------------($6$)
The work done by the locomotive that we are required to find is given by the formula:
$W=F\times s$

This is because work is said to be done by a body when a force is applied in order to move the body by a distance. We now substitute the force and distance given by equation ($5$) and equation ($6$) respectively in the above work equation to get:
$W={\displaystyle \frac{m{a}^2}{2}}\times {\displaystyle \frac{a^2{t}^2}{4}}$
On solving further we get:
$W={\displaystyle \frac{m{a}^4{t}^2}{8}}$
Thus, we have derived the equation for work done by the locomotive. We will now compare this equation with the equation for work done which was already done to find the value for $x$.The work equation given to us is:
$W={\displaystyle \frac{m{a}^4{t}^2}{x}}$
Hence, by equating the two work equations we get:
${\displaystyle \frac{m{a}^4{t}^2}{8}}={\displaystyle \frac{m{a}^4{t}^2}{x}}$
All the common terms cancel out to get the value of $x$. Hence:
$\therefore x=8$

Hence, $8$ is the value of the variable $x$ in the work equation given in the question.

Note: The common mistake that is made in this problem is that the work done equation is not derived again by applying the formula that is known and then comparing the work done equation that is already given and hence this step is the key to finding the value of the required variable.