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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?Quote:
Originally Posted by DINOCALC09
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.
$\displaystyle v^2 = v_0^2 + 2a(y - y_0)$
$\displaystyle v^2 = 2(-9.8)(-0.62)$
$\displaystyle v = -3.48597~m/s$
(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:
$\displaystyle W_{nc} = \Delta E$
$\displaystyle \left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) + (mgy - mgy_0) + \left ( \frac{1}{2}ky^2 - \frac{1}{2}ky_0^2 \right ) = 0$
The ball starts at the top of the spring ($\displaystyle y_0 = 0~m$) with the speed from the last part of the problem, and ends at $\displaystyle y = -0.0445327~m$ with a speed of 0. So:
$\displaystyle -\frac{1}{2}mv_0^2 + mgy + \frac{1}{2}ky^2 = 0$
$\displaystyle k = \frac{\frac{1}{2}mv_0^2 - mgy}{\frac{1}{2}y^2}$
$\displaystyle k = 1421.39~N/m = 14.2139~N/cm$
-Dan
