Question

When a 4kg object is hung vertically on a certain light spring that obeys Hooke’s law, the spring stretches 2.5cm. If the 4kg is removed,

(a) how far will the spring stretch if a 1.5kg block is hung on it?

(b) How much work must an external agent do to stretch the same spring 4cm from its unstretched position?

Short answer

(a) The spring will stretch by 0.9375cm if a 1.5kg block is hung on it.

(b) The work done by the external agent to stretch the spring by 4cm is 1.25J.

Step-by-step solution

Given data

Mass of the first object

$m_1=4\mathrm{kg}$

Mass of the second object

$m_2=1.5\mathrm{kg}$

Displacement of the first object

$\begin{array}{c}x=2.5\mathrm{cm}\\ =2.5·\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)\\ =0.025\mathrm{m}\end{array}$

Stretch caused by the external force

$\begin{array}{c}{x}_e=4\mathrm{cm}\\ =4·\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)\\ =0.04\mathrm{m}\end{array}$

Spring force, work done on a spring and gravitational force

The force that tries to bring a spring to its equilibrium position when it is extended by distance x is

$\mathrm{F}=-\mathrm{kx}$ .....(I)

Here, k is the spring constant.

The work done by a force on a spring of spring constant k when it is extended by x is

$W=\frac{1}{2}k{x}^2$ .....(II)

The downward force on an object of mass m due to gravity is

$\mathrm{F}=\mathrm{mg}$ .....(III)

Here, g is the acceleration due to gravity of value

$g=9.8m/s^2$

Determining the stretch of the 1.5kg block

Equate equations (I) and (III) for the first object to get

$\begin{array}{c}kx=m_{1}g\\ k=\frac{m_{1}g}{x}\end{array}$

Substitute the values to get

$\begin{array}{c}k=\frac{4\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{0.025\mathrm{m}}\\ =1568\mathrm{kg}/{\mathrm{s}}^2\end{array}$

Equate equations (I) and (III) for the second object to get

$\begin{array}{c}kx=m_{2}g\\ x=\frac{m_{1}g}{k}\end{array}$

Substitute the values to get

$\begin{array}{c}x=\frac{1.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{1568\mathrm{kg}/{\mathrm{s}}^2}\\ =0.009375·\left(1\mathrm{m}\times \frac{100\mathrm{cm}}{1\mathrm{m}}\right)\\ =0.9375\mathrm{cm}\end{array}$

Thus, the required stretch is $0.9375\mathrm{cm}$.

Determining the work done by the external force

Calculate equation (II) for the stretch caused by the external force to get

$\begin{array}{c}W=\frac{1}{2}\times 1568\mathrm{kg}/{\mathrm{s}}^2\times {\left(0.04\mathrm{m}\right)}^2\\ =1.25·\left(1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right)\\ =1.25\mathrm{J}\end{array}$

Thus, the required work done is $1.25\mathrm{J}$.