posted by Trace on .
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.
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.