Question

If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.5 cm

Answer 1

Answer:

$\Delta W=24.1162\ J$

Explanation:

Given:

• work done to stretch the spring, $W=4.5\ J$

• length through which the spring is stretched beyond equilibrium, $\Delta x=2.3\ cm=0.023\ m$

• additional stretch in the spring length, $\delta x=3.5\ cm=0.035\ m$

We know the work done in stretching the spring is given as:

$W=\frac{1}{2} \times k.\Delta x^2$

where:

k = stiffness constant

$4.5=0.5\times k\times 0.023^2$

$k=17013.2325\ N.m^{-1}$

Now the work done in stretching the spring from equilibrium to ($\Delta x+\delta x$):

$W'=0.5\times k.(\Delta x+\delta x)^2$

$W'=0.5\times 17013.2325\times 0.058^2$

$W'=28.6162\ J$

So, the amount of extra work done:

$\Delta W=W'-W$

$\Delta W=28.6162-4.5$

$\Delta W=24.1162\ J$