Question

It takes 87 j of work to stretch an ideal spring, initially 1.4 m from equilibrium, to a position 2.9 m from equilibrium. what is the value of the spring constant (force constant) of this spring?

Answer 1

Given:

Work done = 87 Joule

$x_{1}$ = 1.4 m

$x_{2}$ = 2.9 m

To find:

Spring constant = ?

Formula used:

W = $\frac{1}{2} k \triangle x^{2}$

Solution:

Work done required to stretch a spring from 1.4 m to 2.9 m is 87 Joule. Work done is given by following formula.

W = $\frac{1}{2} k \triangle x^{2}$

where W = work done

k = spring constant or force constant

$\triangle x$ is change in length i. e $\triangle x=x_{2} x_{1}$

k = $\frac{2W}{\triangle x^{2}}$

k=$\frac{87 \times 2}{[x_{2} - x_{1} ]^{2}}$

k = 77.33 N/m

Thus, spring constant of the given spring is 77.33 N/m.