Distance a spring has been compressed

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SUMMARY

The discussion focuses on a physics problem involving a 20 kg mass sliding down a frictionless incline and compressing a spring with a spring constant of 200 N/m. The key calculations involve determining the speed of the block just before impact and the distance the spring compresses when the block comes to rest. The correct approach requires accounting for the additional height change as the spring compresses, which affects the total mechanical energy conservation equation. The initial energy is calculated using gravitational potential energy, and the final energy is represented by the spring's potential energy.

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  • Understanding of mechanical energy conservation principles
  • Familiarity with spring potential energy equations (1/2 kx²)
  • Knowledge of gravitational potential energy calculations (mgh)
  • Basic trigonometry, specifically sine and cosine functions
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  • Review the principles of energy conservation in mechanical systems
  • Study the relationship between gravitational potential energy and spring potential energy
  • Learn how to calculate changes in height during spring compression
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Homework Statement



.

A 20 kg mass, released from rest, slides 12 meters down a frictionless plane inclined at an angle of 30º with the horizontal and strikes a spring of spring constant K = 200 Newtons/meter as shown in the diagram above. Assume that the spring is ideal, that the mass of the spring is negligible, and that mechanical energy is conserved. Use g = 10 m/s², (sin30º = ½, cos 30º = 0.866)
a. Determine the speed of the block just before it hits the spring.


b. Determine the distance the spring has been compressed when the block comes to rest.


c. Is the speed of the block a maximum at the instant the block strikes the spring? Justify your answer.



Homework Equations


(1/2)kx^2, (i/2)mv^2


The Attempt at a Solution



For part b, wouldn't I just have to solve

original energy 20*10*12sin(30)= (1/2)(200)x^2

for x?

I do this and I get ~3.46 meters for x, but I'm being told that this is not the right answer.

What am I doing wrong? If energy is conserved, then (1/2)kx^2 would have to equal the original mgh
 
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Joe_I_Am said:
What am I doing wrong? If energy is conserved, then (1/2)kx^2 would have to equal the original mgh
Don't neglect the additional change in height as the spring is compressed. (Measure h from the lowest point, not the point where the mass first touches the uncompressed spring.)
 

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