How far does a block travel before hitting a spring?

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SUMMARY

A block with a mass of 33.0 kg starts from rest on a 31.0-degree inclined plane and compresses a spring with a constant of 3.4 kN/m by 37.0 cm. The potential energy of the spring, calculated as U_s = 232.73 J, is equal to the gravitational potential energy at the top of the incline. The correct approach involves considering the distance traveled before the block encounters the spring, which must include the 37 cm compression. The initial calculation did not account for this distance, leading to an incorrect result.

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  • Understanding of gravitational potential energy (mgy)
  • Knowledge of spring potential energy (U_s = 1/2 kx^2)
  • Trigonometric functions for incline calculations
  • Basic principles of mechanics, including energy conservation
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  • Review energy conservation principles in mechanics
  • Learn about the relationship between spring compression and potential energy
  • Study trigonometric applications in inclined plane problems
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Homework Statement


A block starts from rest at the top of a 31.0 degree inclined plane and encounters a spring, of constant 3.4 kN/m, rigidly attached to the plane. If the block's mass is 33.0 kg and it compresses the spring by 37.0 cm, find the distance the block traveled before it encountered the spring.


Homework Equations





The Attempt at a Solution



The spring compresses to 37cm, and has a spring constant of 3,400 kN. So I solve for U_s, the potential energy of the spring, and get U_s = 232.73 J.

Since friction is not mentioned, I know that the potential energy at the top of the incline equals U_s. Or, mgy = 232.73J, or (33)(9.8)y = 232.73. I use trig and solve for the distance travelled, but it doesn't match the answer of the problem. What am I doing wrong?

Thank you
 
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When you solved for the distanced traveled, did you take into consideration the 37 cm? (They want the distance traveled before hitting the spring.)
 
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That's it. Thank you
 

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