Question

The right end of a relaxed standard spring is at the origin; the left end is clamped at some point on the negative x-axis. Holding the spring’s right end at location x=5 cm requires a force of N. Find the work (in Joules) required to stretch the spring from 5 cm to 8 cm.

Answer 1

The work required to stretch the spring is 350J.

Hooke's Law

This law can be written by the formula: F=kx, where:

F= elastic force (N)

k= spring constant (N/m)

x= linear displacement (m)

Spring Work

For finding the spring work in J, you should apply the formula  $W=\frac{k*(\Delta x)^2}{2}$, where:

W= work (J)

k= spring constant (N/m)

x= difference between the linear displacements (m)

The question gives:

x=5 cm=0.05m requires a force of 1.4N

x=8 cm =0.08m

• Step 1 - First, you should find the spring constant from Hooke's law, for x=5 cm and F=1.4N.

                                          $F=kx\\ 1.4=k*0.05\\ \\ k=\frac{1.4}{0.05} =28\frac{N}{m}$

• Step 2 - Now you can apply the formula for spring work.

                                       $W=\frac{k*(\Delta x)^2}{2}\\ \\ W=\frac{28*(8-3)^2}{2} \\ \\ W=\frac{28*(5)^2}{2}\\ \\ W=\frac{28*25}{2}\\ \\ W=\frac{700}{2}=350J$

Read more about the spring work here:

brainly.com/question/3317535