Extremely Confusing Energy Question - Involves Springs

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SUMMARY

The discussion centers on calculating the maximum compression of a vertical compression spring when a 1.5 kg steel mass is dropped from a height of 0.37 m. The spring has a force constant of 2.1 x 102 N/m. The solution involves equating the gravitational potential energy of the mass to the elastic potential energy stored in the spring. The key equation derived is mg(h+x) = 1/2 kx2, leading to a quadratic equation for maximum compression.

PREREQUISITES
  • Understanding of gravitational potential energy (E = mgy)
  • Familiarity with Hooke's Law (F = kx)
  • Knowledge of elastic potential energy (U = 1/2 kx2)
  • Basic algebra for solving quadratic equations
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn how to derive and solve quadratic equations
  • Explore the applications of Hooke's Law in real-world scenarios
  • Investigate the effects of varying spring constants on compression distances
USEFUL FOR

Students in physics courses, particularly those studying mechanics, engineers working with spring systems, and anyone interested in energy conservation principles.

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I drew out a diagram for this question and wrote out all my givens. I looked through all the formula's I have and I just can't put the pieces of the puzzle together. If anyone can help that would be greatly appreciated.

Question:

A 1.5kg steel mass is dropped onto a vertical compression spring of force constant 2.1 x 10^2 N/m, from a height of 0.37m above the top of the spring. Find, from energy considerations, the maximum distance the spring is compressed.

Thanks!
 
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Work is the integral of force over distance.

The spring applies a force equal to F = kx. The work done in compression is the integral of that force over distance:

W = integral (from 0 to p) (k x dx)
= 1/2 k x^2 | p
= 1/2 k p^2

where p is the maximum compression distance.

Find the gravitational potential energy released by a 1.5 kg mass moving 0.37m downwards (it's just [delta]E = m g y). Set it equal to the work done in the spring's compression, and solve for p.

- Warren
 
I'm guessing that this is not a calculus-based class, is it?

Without calculus it's the same as Chroot said, but it looks a little different. Use conservation of energy, the gravitational potential energy at height "h" above the spring equals the total elastic potential energy of the spring when compressed a distance "x." The trick here is that the distance the mass falls is "h+x." THis is going to lead you to a quadratic solution.
mg(h+x) = 1/2 kx^2.
 

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