Spring constant for ball dropped from 50m height

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lking226
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Homework Statement


A ball is dropped from rest at a height of 50.0m above a spring. After the ball hits, it compresses the spring .340m. Ignoring any non-conservative forces, find the spring constant if the mass of the ball is 4.50 kg.


Homework Equations


Hooke's law: k = mg/x


The Attempt at a Solution


I'm not sure what to do.
k = 4.50 kg x 9.8 m/s / 0.340 m = 129.71
That doesn't seem right.
Help?
 
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but how do you use conservation of energy with springs?
 
lking226 said:
but how do you use conservation of energy with springs?

The potential energy stored in a spring is given by 1/2kx2. So what type of energy is being converted into the elastic potential energy in the spring?
 
the kinetic energy of the ball?
 
lking226 said:
the kinetic energy of the ball?

The energy is possesses at rest 50m above the spring.
 
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?
 
lking226 said:
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?

That would be correct. Can you now form an equation and solve for the spring constant 'k'?
 
mghi + kxi^2 = mghf + kxf^2 ??
 
lking226 said:
mghi + kxi^2 = mghf + kxf^2 ??

Initially the spring is not extended, so what does the left side reduce to? At the final stage, the final height is zero. So the entire equation simplifies to?
 
so then just

mghi = kxf^2
 
lking226 said:
so then just

mghi = kxf^2

Right, so what is k equal to now?
 
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?
 
lking226 said:
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?

That should be

[tex]4.5 kg \times 9.81 m/s^2 \times 50m = \frac{1}{2}k(0.34m)^2[/tex]


Solve again for k
 
To nit-pick, the ball doesn't fall 50.0 meters, it falls 50.34 m after compressing the spring.