Post a New Question

Physics - Re-post

posted by on .

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.

  • Physics - Re-post - ,

    OK: The ball has gravity pulling on, while is supported by something. When that support is moved, the spring holds the ball. The force on the spring is mg, and it gains velocity as it falls, thus gaining KE.

    assume the ball falls h before stopping.
    The energy then in the spring is 1/2 k h^2, and it is equal to the gravity PE that Earth released (mgh).

    mgh=1/2 k h^2
    mg= 1/2 k h

    h= 2mg/k = 2*0.2*9.8/28 m= you do it.

    In my head, I see it about .14m, not even close to .5m

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question