Homework Help: physics

Posted by Trace on Wednesday, November 24, 2010 at 12:22am.

A mass of 1.00kg is dropped from a height of 80cm onto a spring, of unstretched length of 20.0cm, which is resting on the floor. It comes to rest, momentarily, at a height of 10.0cm from the floor. If we can ignore friction, find the spring constant of the spring.

I feel like spring questions are so confusing. I'm not even sure what to do. I calculated -Wgravity=mgDELTAy= 7.84J

Now, can I equate -Wgravity=Uelastic=1/2kx^2?

I'm so confused.

physics - drwls, Wednesday, November 24, 2010 at 4:02am
Stretching the spring 10 cm casues it to have a spring potential energy equal to the gravitational potential energy loss.

(1/2) k*(0.10m)^2 = M g H = 1.0*9.8*0.7
= 6.86 J
Solve for k

0.7 m (70 cm) is the actual distance it has fallen at maximum compression.
0.10 m is the spring comoression.

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