What is the force constant and work done in stretching a Hooke's-law spring?

[jrl3253]

Homework Statement

The force required to stretch a Hooke's-law spring varies from 0 N to 70.7 N as we stretch the spring by moving one end 6.52 cm from its unstresses position. Find the force constant of the spring. answer in units of N/m.
Find the work done in the stretching the spring. Answer in units of J.
F=0-70.7 N
d= 6.52 cm

Homework Equations

W=Fd
W=-Kx

The Attempt at a Solution

W=70.7*6.52=460.964

 

[JesseC]

what is hooke's law? you haven't written that down and it would be the most relevant equation to find the force constant of the spring!

I would add that once you have found the force constant from hookes law, then you can use it to work out the work done. Due to conservation of energy, work done stretching a spring is equal to elastic (potential) energy stored in the spring.

do you know the equation for the elastic energy in a spring? that's the other relevant equation!

 

[jrl3253]

hooke's law is in the post...W=-kx

 

[JesseC]

hooke's law is in the post...F=-kx

2. Homework Equations

W=Fd
W-Kx

I don't see it there!

In your attempt at a solution, you skipped straight to finding W (Work). You cannot calculate this without first knowing k (force constant).

Don't forget that distances should be in meters for calculations if you want to get the right answer in joules.

 

[rwisz]

F=70.7 N
d=6.52 cm=0.0652 m

To find the force constant, we can rearrange the equation W=-Kx to solve for K:
K = -W/x = -460.964 J / 0.0652 m = -7060.06 N/m

Therefore, the force constant of the spring is 7060.06 N/m.

To find the work done in stretching the spring, we can use the equation W=Fd with F=70.7 N and d=0.0652 m:
W=70.7 N * 0.0652 m = 4.611 J

Therefore, the work done in stretching the spring is 4.611 J.