Asked by Jennifer

A .20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball is supported initially so that the spring is neither stretched or compressed, is released from rest. How far does the ball fall before brought to a momentary stop by the spring?

Our teacher gave us a hint:
PE+KE=PE+KE

But I think that there is something missing, perhaps subscripts (initial and final). And if that's the case, then, I'll get PE=PE.

Also, someone else in class explained that PE=mgh and solved the problem, but I do not know how she got .5 for h.

Please help, so confused. Thank you.
Answered by Massy
ETi = ETf
Ek+ Eg+Ee = Ek+Eg+Ee
since vi and vf is o Ek is taken out, and having a ref point at the bottom Eg is for ETf is taken out( final hight =0 meters). Also spring is at equilibirum in the beinging, so Ee for ETi is also taken out( xi = O meters)
so then Eg= Ee
and mghi= 0.5(k)(xf)^2
ur hi= xf
mghi= 0.5(k)(hi)^2
hi=mg/0.5(k)
hi= (0.20)(9.80)/(0.5)(28)
hi= 0.14

wats weird is my gr.12 physics teacher gave the same exact question, kk hope it helped! good luck
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