Conservation of Energy Problem-I think

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SUMMARY

The discussion centers on a conservation of energy problem involving a freight car rolling down a 2.8 m incline and compressing a spring with a constant of 4.9 x 106 N/m. The initial approach incorrectly calculated the spring compression as 0.255 m, but after correcting for gravitational acceleration (g), the accurate compression was determined to be 0.8 m. The key equations utilized include the conservation of energy formula and the relationship between potential energy and spring compression. The final solution emphasizes the importance of including gravitational force in energy calculations.

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  • Understanding of conservation of energy principles
  • Familiarity with potential energy (PE) and kinetic energy (KE) equations
  • Knowledge of spring mechanics and Hooke's Law
  • Basic algebra for solving equations involving square roots
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Conservation of Energy Problem--I think

In a switchyard, freight cars start from rest and roll down a 2.8 m incline and come to rest against a spring bumper at the end of the track (Fig. 8-27). If the spring constant is 4.9 x 106 N/m, how much is the spring compressed when hit by a 57,000 kg freight car?

I thought that this was a conservation of energy problem, so I set up K_i + U_si + U_gi = K_f + U_sf + U_gf, and after cancelling out all the things that equal zero, I got PE_gi = PE_sf. Then I did mgy_i = .5(k)(x)^2. When I solved for change in x, I got 0.255 m, but that was wrong, so I don't know where to go from there. Any help would be greatly appreciated. Thanks! Beverly
 
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bbbbbev said:
In a switchyard, freight cars start from rest and roll down a 2.8 m incline and come to rest against a spring bumper at the end of the track (Fig. 8-27). If the spring constant is 4.9 x 106 N/m, how much is the spring compressed when hit by a 57,000 kg freight car?

I thought that this was a conservation of energy problem, so I set up K_i + U_si + U_gi = K_f + U_sf + U_gf, and after cancelling out all the things that equal zero, I got PE_gi = PE_sf. Then I did mgy_i = .5(k)(x)^2. When I solved for change in x, I got 0.255 m, but that was wrong, so I don't know where to go from there. Any help would be greatly appreciated. Thanks! Beverly

PE_final + KE_final = PE_initial + KE_initial
0 + KE_final = (57000)*g*2.8 + 0

KE_final = (1/2)mv^2 = 0.5kx^2 = 57000*g*2.8
x = root (57000*g*2.8 / (0.5*4.9*10^6)) = 0.8 meters?
 
Thanks! I had forgotten to multiply by g somehow! I can't believe I didn't notice that before!
 

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