Question

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the spring, in turn, as shown in Fig.${}^{\mathbf{7}\mathbf{-}\mathbf{48}}$. (a) Which mark on the scale will the pointer indicate when no package is hung from the spring? (b) What is the weight W of the third package?

Short answer

a. The mark on the scale as indicated by the pointer when no weight is hung from the spring i${}^{\mathrm{x}=23\mathrm{mm}}$.

b. Weight of the third package is ${}^{45\mathrm{N}}$.

Step-by-step solution

Given information

The figure for three mass-spring systems are given

Determining the concept

The problem is based on the Hooke’s law. It states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively modest deformations of an object. Find the mark on the scale as indicated by the pointer when no weight is hung from the spring by solving the two equations got after applying Hooke’s law to system 1 and 2.Then from this, find the spring constant. Then applying Hooke’s law to the third system, find the weight of the third package.

Formula:

Force by the Hooke’s law is given by,

$\mathrm{F}=-\mathrm{k}(\mathrm{x}-{\mathrm{x}}_{0)}$

Where ,F is force, x,${}^{{\mathrm{x}}_0}$are displacements and k is the spring constant.

(a) Determining the mark on the scale as indicated by the pointer when no weight is hung from the spring

Applying Hooke’s law to the first and second system yields,

$$\begin{gathered} -110=-\mathrm{k}(40-{\mathrm{x}}_0) \\ -\mathrm{k}=-\frac{110}{40-{\mathrm{x}}_0}\left(\mathrm{i}\right) \\ -240=-\mathrm{k}(60-{\mathrm{x}}_0) \\ -\mathrm{k}=-\frac{240}{60-{\mathrm{x}}_0}\left(\mathrm{ii}\right) \\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right), \\ -\frac{240}{60-{\mathrm{x}}_0}=-\frac{110}{40-{\mathrm{x}}_0} \\ 9600-240{\mathrm{x}}_0=6600-110{\mathrm{x}}_0 \\ 130{\mathrm{x}}_0=3000 \\ {\mathrm{x}}_0=23\mathrm{mm} \\ \mathrm{Hence},{\mathrm{x}}_0=23\mathrm{mm}\mathrm{is}\mathrm{the}\mathrm{mark}\mathrm{of}\mathrm{the}\mathrm{scale}\mathrm{as}\mathrm{indicated}\mathrm{by}\mathrm{the}\mathrm{pointer}\mathrm{when}\mathrm{no}\mathrm{weight} \\ \mathrm{is}\mathrm{hung}\mathrm{from}\mathrm{the}\mathrm{spring}. \end{gathered}$$

(b) determining the weight of the third package

Putting value of in equation (i) gives,

$$\begin{gathered} -\mathrm{k}=-\frac{110}{40-23} \\ \mathrm{k}=6.47 \\ ~6.5\mathrm{N}/\mathrm{m} \\ \mathrm{Applying}\mathrm{Hooke}\text{'}\mathrm{s}\mathrm{law}\mathrm{to}\mathrm{the}\mathrm{third}\mathrm{system}\mathrm{yields}, \\ \mathrm{W}=\mathrm{k}(30-{\mathrm{x}}_0) \\ \mathrm{W}=6.5(30-23) \\ \mathrm{W}=45\mathrm{N} \\ \mathrm{Therefore},\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{third}\mathrm{package}\mathrm{is}45\mathrm{N}. \end{gathered}$$