Conservation of energy up a ramp.

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SUMMARY

The discussion centers on the conservation of energy involving a block attached to a spring with spring constant k, compressed a distance x_1. When released, the block slides up a ramp, reaching a height h_1 determined by the equation h_1 = (k x_1^2) / (2mg). When the spring is compressed to 2x_1, the maximum height h_2 is calculated as h_2 = (k (2 x_1)^2) / (2mg), leading to the conclusion that h_2 is four times h_1. The confusion arises from a differing answer suggesting a height increase of only √2 times, prompting a reevaluation of the problem's parameters.

PREREQUISITES
  • Understanding of spring potential energy (SPE) and gravitational potential energy (GPE)
  • Familiarity with the conservation of energy principle
  • Basic algebra for manipulating equations
  • Knowledge of the relationship between force, mass, and acceleration due to gravity (g)
NEXT STEPS
  • Review the principles of conservation of energy in mechanical systems
  • Study the effects of varying spring constants on energy transfer
  • Explore the implications of frictionless surfaces in energy conservation problems
  • Investigate the relationship between compression distance and potential energy in springs
USEFUL FOR

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

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Homework Statement


Suppose there is a block attached to a spring with spring constant k. The block is pushed so that it compresses the spring a distance x_1. The block is released and slides without friction up a ramp, coming to a maximum height h_1 above the ground. Suppose we compressed the spring twice as far, what maximum height does the block now reach?

Homework Equations


SPE = \frac{k x^2}{2}
GPE = m g h

The Attempt at a Solution


Conservation of energy, for the first situation then the second:
SPE=GPE
\frac{k x_1^2}{2} = mgh_1
h_1=\frac{k x_1^2}{2mg}

SPE=GPE
\frac{k (2 x_1)^2}{2} = mgh_2
h_2=\frac{k 4x_1^2}{2mg}

\frac{h_2}{h_1} =\frac{k 4 x_1^2}{2mg} * \frac{2mg}{k x_1^2} = 4

However the answerI have (this is from a course I took a long time ago) says it goes \sqrt{2} times as high. Did I do something stupid here?
 
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I agree with your result.
Maybe the original question was the other way round? If the height should double, which factor do you need for the spring?
 

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