Block compresses spring, find coefficient of kinetic friction

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a 2.5 kg block released from a compressed spring with a spring constant of 200 N/m. The block compresses the spring by 18 cm and slides 90 cm before stopping. The elastic potential energy stored in the spring is calculated as 7.29 J, while the normal force acting on the block is determined to be 24.5 N. The relationship between the initial mechanical energy, the change in thermal energy due to friction, and the final mechanical energy is emphasized as crucial for solving the problem.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of elastic potential energy calculations
  • Familiarity with the concepts of kinetic energy and friction
  • Basic principles of mechanical energy conservation
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  • Calculate the coefficient of kinetic friction using the formula: μk = Fk / Fn
  • Explore energy conservation principles in mechanical systems
  • Investigate the effects of varying spring constants on kinetic energy
  • Learn about the relationship between work done by friction and energy loss
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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
 
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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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