Solve Spring Compression Homework Problem

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SUMMARY

The discussion focuses on solving a physics problem involving a spring and a block. The block, weighing 1.65 kg, moves at 5.58 m/s before compressing the spring, which is fixed at one end. The coefficient of kinetic friction between the block and the table is 0.300. The key takeaway is that energy conservation principles must be applied, where the initial kinetic energy of the block is reduced by the work done against friction and the energy stored in the compressed spring.

PREREQUISITES
  • Understanding of kinetic energy calculations (KE = 0.5 * mass * velocity²)
  • Knowledge of friction force calculations (F of kinetic friction = coefficient of kinetic friction * mass * g)
  • Familiarity with the concept of energy conservation in physics
  • Basic understanding of spring mechanics and Hooke's Law
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about Hooke's Law and how it applies to spring compression
  • Explore frictional forces and their impact on motion
  • Practice solving similar physics problems involving springs and kinetic energy
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of spring systems and energy conservation in mechanics.

andrewn
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Homework Statement


Could somebody help walk me through this problem?:
A spring has its right end fixed and is installed on a horizontal table so that the free end, in equilibrium, is at x= 3m. A 1.65 kg block coming from the left slides along the table. When it passes the origin, it is moving at 5.58 m/s. It strikes the spring, compresses it momentarily, and is then sent back toward the left, where it eventually comes to rest at the point x= 1.5m. The coefficient of kinetic friction betweeen the block and the table is .300. By what distance was the spring compressed?


Homework Equations


F(of kinetiic friction)=(coefficeint of kinetic friction)(m)(g)
KE=.5(mass)(velocity squared)

The Attempt at a Solution


KE=.5(1.65kg)(5.58 m/s squared)
=25.57Joules
F(of kinetic friction)=.3(1.65)(9.8)
=4.851 N
 
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You seem to have ended up with a force not a distance!

You had almost the correct approach, energy is conserved
First it is the KE of the incoming block.
Then some is lost in force*distance of the friction on the table.
Then it is the force*distance of the compressed spring (when the block is at rest) then it is the force*distance the retreating block travels.
 
I'd use this idea... Work done by friction = final energy - initial energy

The distance traveled into the spring is a variable in the left side of the equation.
 

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