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Nick-
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You push a 2.5 kg block against a horizontal spring, compressing the spring by 18 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 90 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?
Having trouble finding the initial and final mech energy (sum of kinetic and elastic potential energy - I don't know how to find the velocity so I can't find the kinetic energy, would you set up energy equations?)
Using the spring constant I found: F=-k(d) = -200N/m (.18m) = 36 N
Elastic potential energy = 1/2kx^2 = 1/2(450N/m)(.18m)^2 = 7.29 N
The change in thermal energy = Fk(d) = (Uk)(Fn)(d)
Normal Force = mg = 24.5 N
Also, initial mech energy - change in thermal energy = final mech energy
That's pretty much all I got, any help would be much appreciated, thanks
Having trouble finding the initial and final mech energy (sum of kinetic and elastic potential energy - I don't know how to find the velocity so I can't find the kinetic energy, would you set up energy equations?)
Using the spring constant I found: F=-k(d) = -200N/m (.18m) = 36 N
Elastic potential energy = 1/2kx^2 = 1/2(450N/m)(.18m)^2 = 7.29 N
The change in thermal energy = Fk(d) = (Uk)(Fn)(d)
Normal Force = mg = 24.5 N
Also, initial mech energy - change in thermal energy = final mech energy
That's pretty much all I got, any help would be much appreciated, thanks