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K in Hookes Law vs. Spring Potential Energy

Today in physics class we were discussing the conservation of energy using a ball on a spring as the example. When the instructor completed the problem one of the students stated that the value of K (spring constant) that we found was different (by a factor of 2) than the value of K calculated using Hooke's Law. The instructor could not figure out why the values of K were not the same. Why are they different?

Potential Energy(before)=Potential Energy (after)
(1/2)kx^2+mgh = (1/2)kx^2+mgh
ball on spring at rest = ball on spring extended

2mgh
---------- = k is not equal to k= F/delta h
X^2
 

lightgrav

Homework Helper
1,248
30
You can equate the sum of these Potential Energies only if KE = 0.
This will occur at the bottom of the bounce and at the top of the bounce.
[tex]\frac{1}{2} k x_{bottom}^2 + m g h_{bottom} = \frac{1}{2} k x_{top}^2 + m g h_{top}[/tex]
suppose the spring was unstretched at the top, so x_top = 0 ;
suppose we measure height from the bottom , so h_bottom = 0 .
Then : (1/2) k (x_bottom)^2 = m g h_top
=> k = 2 m g h_top / (x_bottom)^2 = 2 m g / x_bottom .

The adjectives "top" and "bottom" are important here. The ball was moving
- in particular, the acceleration was NOT = 0 in either location.

The Force by the spring when the ball was at the bottom was
F = - k s = - [2 m g / x_bottom] x_bottom = - 2 m g (upward).

The Force by the spring when the ball was at the top was zero.
The average Force by spring during this motion was F_average = - m g .

The form to remember is F = - k s , which will remind you that
s is the stetch of the spring from its "relaxed" length.
(not to be confused with horizontal coordinate "x" , nor with height)
 

Hootenanny

Staff Emeritus
Science Advisor
Gold Member
9,598
6
There are two versions of Hooke's law:
[tex]T = \frac{kx}{L}[/tex]

[tex]F = kx[/tex]
Each equation will yield a different value of k for a given tension or (after integration) potential energy.
 

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