Block compresses spring, find coefficient of kinetic friction

AI Thread Summary
To find the coefficient of kinetic friction for the block sliding on the tabletop after being released from the spring, the initial elastic potential energy of the spring must be calculated using the formula 1/2kx^2, resulting in 7.29 J. The block, with a mass of 2.5 kg, experiences a normal force of 24.5 N, and the work done against friction can be expressed as Fk(d), where d is the distance the block slides (90 cm). The force exerted by the spring (36 N) translates into work that equals the change in mechanical energy, allowing for the calculation of the frictional force. By equating the initial mechanical energy to the work done against friction, the coefficient of kinetic friction can be determined. The discussion emphasizes the relationship between potential energy, work done, and kinetic friction in this scenario.
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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
 
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Nick- said:
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

Welcome to PF.

You know how much potential energy is in the spring just before it's released.

So what force acting over the 90 cm distance will have absorbed the PE that went into the object's KE? Won't that work equal the Potential it had when it was on the spring?
 
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