A 52.5 kg mass attached to a spring oscillates with an amplitude of 1.95 m. The total mechanical energy of the system is 395 J. Find the position of the mass when its speed is 2.00 m/s.

Question

anonymous · Answer

We know that the total mechanical energy of the mass-spring system is defined as: $$E = {1 \over 2} m v^2 + {1 \over 2} kx$$which is the sum of the kinetic and potential energies of the system. We don't know the spring constant, $k$. This can be found from the fact that we know the mass will have a zero velocity at the maximum amplitude. Therefore, $$395 = {1 \over 2}k x ~~~ @ x = 1.95$$After solving for $k$, we can use the following expression to find the position where the velocity is equal to $\rm 2~ {m \over s}$$$395 = {1 \over 2} m (2) + {1 \over 2} k x$$We can solve for x.

anonymous · Answer

I got that k = 405.128
then I plugged it in, and I got x=1.43m. Is that right? 
THANK YOU SO MUCH!!!

underhill · Answer

Eashmore's first equation is wrong.  The potential energy of a spring system is$$1/2(k)(x^2)$$

anonymous · Answer

Whoops! Thanks for the catch Underhill.