spring force

Quote:

Originally Posted by DINOCALC09

A 135 g ball is dropped from a height of 62 cm above a spring of negligible mass. The ball compresses the spring to a maximim displacement of 4.45327 cm. The acceleration of gravity is 0.8 m/s^2.

Calculate the spring force constant k. (units are N/m>

I think you meant "9.8 m/s^2" yes?

For simplicity take this problem in two parts. (You can do this in one step, but it's messy.)

First part: The ball falls to the level of the top of the spring (ie. falls 62 cm).

How fast is it moving just before it hits the spring? I've got a coordinate system at the point where the ball is dropped and I'm calling +y upward.
$v^2 = v_0^2 + 2a(y - y_0)$

$v^2 = 2(-9.8)(-0.62)$

$v = -3.48597~m/s$
(Negative since it's falling downward.)

Second part: The ball compresses the spring.

This is a work - energy problem. I'm defining the 0 level for the gravitational potential energy to be at the top of the uncompressed spring, and the 0 level for the spring potential energy to be at the top of the uncompressed spring, and in both cases I'm taking +y upward. There are no non-conservative forces at work here, so:
$W_{nc} = \Delta E$

$\left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) + (mgy - mgy_0) + \left ( \frac{1}{2}ky^2 - \frac{1}{2}ky_0^2 \right ) = 0$

The ball starts at the top of the spring ( $y_0 = 0~m$ ) with the speed from the last part of the problem, and ends at $y = -0.0445327~m$ with a speed of 0. So:
$-\frac{1}{2}mv_0^2 + mgy + \frac{1}{2}ky^2 = 0$

$k = \frac{\frac{1}{2}mv_0^2 - mgy}{\frac{1}{2}y^2}$

$k = 1421.39~N/m = 14.2139~N/cm$

-Dan