Why Do Spring Constants Vary With Different Calculation Methods?

[mike115]

[SOLVED] Spring - Forces or Energy?

Homework Statement

A spring of length 0.80 m rests along a frictionless 30 degree incline. A 2.0 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m.
A) Determine the spring constant k .
B) The mass is pushed down, compressing the spring an additional 0.60 m, and then released. If the incline is 2.0m long, determine how far beyond the rightmost edge of the incline the mass lands.

Homework Equations

Fspr = k*x
Wspr = .5*k*x^2

The Attempt at a Solution

For part A, I get two different answers for the spring constant depending on whether I try to use forces or energy. Can somebody explain why one way is correct? Why doesn't forces/energy give the correct answer?

Fspr = Fg parallel to the incline
k*x = m*g*sin(30)
k*.1 = 2*9.8*sin(30)
k = 98 N/m

Wspr = PEg
.5*k*x^2 = m*g*x*sin(30)
.5*k*.1^2 = 2*9.8*.1*sin(30)
k = 196 N/m

Thanks.

 

[olgranpappy]

you can't blindly use those equations for potential energy. you have to think about what you are doing. there would be a kinetic contribution if the box were just to drop down on the spring.

as it is you can suppose that someone slowly let the box set down and compress the spring. the spring did work on the box etc etc, but who cares.

use the fact that the box is already just sitting there (regardless of how the box got there) at rest (not accelerating either) so that the forces must be equal.

 

[asura]

can you explain how to do part B

im not exactly sure how to set it up
heres what i have:

(mass = 2.4 kg and k = 117.72 N/m)

(PEg + PEs)i = (PEg + KE)f
mgh + .5kx^2 = mgh + .5mv^2

2.4 kg (9.81 m/s^2) (0.1 m) sin 30 + .5(117.72 N/m)(0.60 m)^2 = 2.4 kg (9.81m/s^2)(0.70 m) sin 30 + .5(2.4 kg) v^2

what I am most confused about is where I should consider "final" to find velocity
so for example is it when the spring is completely relaxed or is it when the mass is resting on it (compressing it .1 m)