Question

The propelling force of a rocket increases uniformly from zero to 5 N in the first 12 m and remains constant for the next 40 m. Find the total work done.

Answer 1

Step 1: Given data

From a distance of $x=0m$ to $x=12m$, force changes from zero to, $F=5N$

Force remains constant for the next, $d=40m$

Step 2: Calculating proportionality constant

Using the Hooke's law, it can be said that,

$F\propto x$

where, $F$ is force and $x$ is distance.

Therefore,

$F=kx$

where, $k$ is a proportionality constant.

Substituting $F=5N$ and $x=12m$ in the above equation, we get,

$$\begin{gathered} 5N=k\times 12m \\ k=\frac{5}{12} \end{gathered}$$

Step 3: Calculating work done in the first 12 m

Let the work done in the first $12m$ be $W_1$. Therefore, work done from a distance of $x=0m$ to $x=12m$ can be calculated as follows:

$$\begin{gathered} W_1={\int }_0^{12}Fdx \\ ={\int }_0^{12}kxdx \\ =k{\int }_0^{12}xdx \\ =\frac{5}{12}\times {\left[\frac{x^2}{2}\right]}_0^{12} \\ =\frac{5}{12}\times \left[\frac{{12}^2}{2}-\frac{0^2}{2}\right] \\ =\frac{5}{12}\times 72 \\ =30J \end{gathered}$$

Step 4: Calculating work done for the next 40 m

Work done is given as,

$W=Fd$

where, $F$ is force and $d$ is displacement.

Let work done in the next $40m$ be $W_2$.

Now for $d=40m$, force is $F=5N$(Since force remains constant)

Therefore, work done will be,

$$\begin{gathered} W_2=5N\times 40m \\ {W}_2=200J \end{gathered}$$

Step 5: Calculating total work done

The total work done will be the work done in the first $12m$ and the work done in the next $40m$. Therefore, the total work done will be,

$$\begin{gathered} W=W_1+W_2 \\ W=30J+200J \\ W=230J \end{gathered}$$

Therefore, total work done is $230J$.