# physics

In preparation for shooting a ball in a pinball machine,a spring (K=675N/m) is compressed by 0.0690m relative to its unstrained length. The ball(m=0.600kg) is at rest against the spring at point A.When the spring is released, the ball slides (without rolling) to point B, which is 0.300m higher than point A. How fast is the ball moving at B?
solution:Ef=E0
1/2 MVsq+1/2kXfsq =1/2kx0sq
vf= sqrootk/m(Xintial sq+xfinal sq)
vf= [675/0.600kg(0.690m+.300m)]sqroot
vf=20.37m/s2

I don't think it is right

No. Your algebra is in error (sign).

I get something like..

v2=k/m (xf2 - xo2 )
then solve for v.

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