Spring Compression on an incline

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SUMMARY

The discussion focuses on calculating the maximum compression of a spring when an 11 kg box slides down a frictionless ramp inclined at 30 degrees. The spring constant is 190 N/m. The maximum compression of the spring is determined to be 0.56795 m using the energy conservation principle, where the initial gravitational potential energy is converted into elastic potential energy. The equation used is PEgravity initial = PEspring final, simplifying to 11 * 9.81 * sin(30) * 4 = 1/2 * 190 * (Δx)^2.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgh)
  • Familiarity with elastic potential energy (PE = 1/2 * k * (Δx)^2)
  • Knowledge of energy conservation principles in physics
  • Basic trigonometry, specifically sine functions in relation to angles
NEXT STEPS
  • Explore the concept of energy conservation in mechanical systems
  • Learn about the dynamics of springs and Hooke's Law
  • Investigate the effects of friction on inclined planes
  • Study the relationship between potential energy and kinetic energy in motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to explain spring dynamics and energy transformations.

luvey
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Homework Statement


A 11kg box slides 4.0m down the frictionless ramp which has an incline of 30o to the horizontal, then collides with a spring whose spring constant is 190 N/m.

a.What is the maximum compression of the spring?
b.At what compression of the spring does the box have its maximum speed?


Homework Equations


Spring: PE=1/2*k([tex]\Delta[/tex]s)2
Gravity: PE=mgsin([tex]\theta[/tex])d
PEspring initial+PEgravity initial=PEspring final+PEgravity final


The Attempt at a Solution


I tried plugging in the numbers which gave me:
0+11*9.81*sin(30)*4=1/2*190([tex]\Delta[/tex]s)2+11*9.81*sin(30)*(4+[tex]\Delta[/tex]s)
Which gave me the max compression of 0.56795 m

I haven't made it to part b yet because I can't seem to obtain the max compression.
 
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luvey said:

The Attempt at a Solution


I tried plugging in the numbers which gave me:
0+11*9.81*sin(30)*4=1/2*190([tex]\Delta[/tex]s)2+11*9.81*sin(30)*(4+[tex]\Delta[/tex]s)
Which gave me the max compression of 0.56795 m

I haven't made it to part b yet because I can't seem to obtain the max compression.

You do not need the extra 11*9.81*sin(30) on the right side.

The left side is the initial energy which when it reaches the spring is being entirely stored as elastic potential energy.
 

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