Question

A force is applied to an ideal spring (initially in its equilibrium position) and does 1.9 JJ of work stretching it 2.2 cmcm . How much work is required to stretch the spring by 6.5 cmcm from its equilibrium position?

Answer 1

Answer:

W=16.58J

Explanation:

initial information we have

work: $W=1.9J$

stretched distance: $x=2.2cm=0.022m$

from this, we can find the value of the constant of the spring k, with the equation for work in a spring:

$W=\frac{1}{2} kx^2$

substituting known values:

$1.9J=\frac{1}{2}k(0.022)^2$

and clearing for k:

$k=\frac{2(1.9J)}{0.022^2} \\k=7,851.24$

and now we want to know how much work is done when we stretch the spring a distance of 6.5cm from equilibrium, so now x is:

$x=6.5cm=0.065m$

and using the same formula for work, with the value of k that we just found:

$W=\frac{1}{2} kx^2$

$W=\frac{1}{2}(7851.24)(0.065)^2\\W=16.58J$

Answer 2

Answer:

16.586 J

Explanation:

The energy stored in an elastic material is given as,

E = 1/2ke².................. Equation 1

Where E = work done by the  ideal spring, k = force constant of the spring, e = extension.

make k the subject of the equation

k = 2E/e²..................... Equation 2

Given: E = 1.9 J, e = 6.5 cm = 0.022 m

Substitute into equation 2

k = 2(1.9)/0.022²

k = 3.8/(0.022²)

k = 7851.24 N/m

When the spring is stretched by 6.5 cm

E = 1/2ke'²................. Equation 3

Where e = 6.5 cm = 0.065

Substitute into equation 3

E = 1/2(7851.24)(0.065²)

E = 16.586 J.