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?
Find the stored energy in the spring: 1/2 k x^2
Then conservation of energy states...
1/2 kx^2 = 1/2 mass*vb^2 + mg(.300+.0690)
solve for vb
check my thinking.
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
1 answer