How Does Hooke's Law Determine Spring Constant and Extension?

[mandy9008]

Homework Statement

Hooke's law describes a certain light spring of unstretched length 33.0 cm. When one end is attached to the top of a door frame, and a 5.80-kg object is hung from the other end, the length of the spring is 42.50 cm.
(a) Find its spring constant.
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 170 N. Find the length of the spring in this situation.

Homework Equations

F=ma
F=-kx

The Attempt at a Solution

a. F=ma
F=(5.8kg)(9.8m/s²)
F=56.84N

F=-kx
56.84N = -k (.425m - .330m)
k=598.3 N/m or 0.598 kN/m

b. F=-kx
170N = -598.3 N/m x
x= 284.3m

 

[wwshr87]

I did the first part using work done by the mass and the spring and got a spring constant of 1196.63 N/m. For part b since each is pulling I believe you should sum both forces, good luck.

 

[inky]

a. F=ma
F=(5.8kg)(9.8m/s²)
F=56.84N

F=-kx
56.84N = -k (.425m - .330m)
k=598.3 N/m or 0.598 kN/m

b. F=-kx
170N = -598.3 N/m x
x= 284.3m[/QUOTE]


I think part (a) is correct. Part (b) is in correct. Two applied forces are same magnitude and opposite directions. It will cancel each other. You need to consider weight of the object. So extension is previous extension.