13.7 Work by gravity and spring force (check your understanding)

Speeds at initial and final poistions are same. Therefore, total work done by gravity and friction during downward motion is zero according to "work-kinetic energy" theorem. It means that work done during downward motion by gravity (mgh) equals work done by friction (-mgh), which is opposite in sign. During upward motion, both gravity and friction oppose motion of the object. Hence, work by external force must equal to the magnitude of negative works by gravity and friction.

$$\begin{array}{l}{W}_{\mathrm{Ext}}=W_{\mathrm{Gravity}}+W_{\mathrm{Friction}}\\ \Rightarrow W_{\mathrm{Ext}}=2W_{\mathrm{Gravity}}\\ \Rightarrow W_{\mathrm{Ext}}=2mgh\end{array}$$

Hence, option (b) is correct.

Work done by the external force in stretching spring by "x" is :

$$\begin{array}{l}W=\frac{1}{2}k{x}^2\end{array}$$

We need to eliminate extension "x" in terms of "F" and "k" in order to make the comparison for same force. Now, external force on the spring is :

$$\begin{array}{l}F=kx\end{array}$$

Combining two equations, we have :

$$\begin{array}{l}W=\frac{F^2}{2k}\end{array}$$

For same external force,

$$\begin{array}{l}\Rightarrow \frac{W_1}{W_2}=\frac{k_2}{k_1}\end{array}$$

As $k_1>k_2$ ,

$$\begin{array}{l}\Rightarrow W_1<W_2\end{array}$$

Hence, option (b) is correct.

We need to obtain an expression for the speed "v". This relation can be obtained, using "work - kinetic energy" theorem :

$$\begin{array}{l}\frac{1}{2}m{v}^2=mgl\sin \theta -{\mu }_{k}mgl\cos \theta \\ \Rightarrow \frac{1}{2}{v}^2=gl\sin \theta -{\mu }_{k}gl\cos \theta \end{array}$$

Clearly, the speed at the bottom of the incline is independent of mass "m".

Hence, option (a) is correct.

The work done by a spring force for an extension "x" is :

$$\begin{array}{l}{W}_S=-\frac{1}{2}k{x}^2\end{array}$$

However, spring does this work on blocks of same mass, which are pulled by equal horizontal force. The work by spring is, thus, equally divided on two blocks. Hence, work done by the sping on each block is :

$$\begin{array}{l}{\mathrm{W\text{'}}}_S=-\frac{1}{4}k{x}^2\end{array}$$

Hence, option (d) is correct.

The first part of the question can be used to determine the spring constant. The work done by external force is :

$$\begin{array}{l}W=\frac{1}{2}k{x}^2\end{array}$$

$$\begin{array}{l}\Rightarrow k=\frac{2W}{x^2}\\ \Rightarrow k=\frac{2x4}{{0.1}^2}=800N/m\end{array}$$

Now, work done to extend the spring by a total of 20 cm,

$$\begin{array}{l}\Rightarrow W=\frac{1}{2}x800x{0.2}^2=16J\end{array}$$

Thus, additional work required is 16 - 4 = 12 J

Hence, option (c) is correct.

Let "v" be the terminal speed. The change in kinetic energy ( $\frac{1}{2}m{v}^2$ ) is equal to the work done by gravity and viscous force. Gravity does the positive work, whereas viscous force does negative force. When the drop attains terminal velocity, then viscous force equals the weight of the rain drop,

$$\begin{array}{l}F=mg\end{array}$$

$$\begin{array}{l}\Rightarrow 6\pi \eta rv=mg\end{array}$$

$$\begin{array}{l}\Rightarrow v=\frac{mg}{6\pi \eta r}\\ \Rightarrow v=\frac{\frac{4}{3}\pi r^3\rho g}{6\pi \eta r}\\ \Rightarrow v\propto r^2\\ \Rightarrow v=A{r}^2\end{array}$$

where "A" is a constant. Now, work done by the forces :

$$\begin{array}{l}W=\frac{1}{2}m{v}^2\\ \Rightarrow W=\frac{1}{2}x\frac{4}{3}\pi r^3\rho A^2{r}^4\\ \Rightarrow W\propto r^7\end{array}$$

Hence, option (d) is correct.