A certain spring has a spring constant of 165 N/m. A 20kg mass is attached to the end of the spring. While the mass is at the equilibrium position, it is given a velocity of 8m/s downward.

Question

What is the total energy of the system, and aacceleration of the system when it is at the classical turning point? Also, what is the maximum distance that the spring is stretched? Lastly, what must be the displacement,x, be so that the acceleration of the mass is 9.8 m/s^2. I know that i have to use ME=EPE+KE. EPE=1/2kx^2 and KE=1/2mv^2. I just don't know how to apply them here.

Answer 1

<vx)-ln[.03]/(l1)<l3>
A shortcut I forgot about until calc III is to make it a function of <vx)
vx=nnx+bbn+c
c(x)=<l.5/.556>
nn(x)
x=nn(.565)
bbn
n=(bb)=.909
c=vvx/.909[pi]^2<nnx>

TOTAL DISPLACEMENT= ____