Conservation of Energy Problem-I think

AI Thread Summary
The discussion centers on a conservation of energy problem involving a freight car rolling down an incline and compressing a spring. The initial setup involved equating potential energy to spring energy, but the user initially miscalculated by not including gravitational acceleration. After correcting the calculation by incorporating the gravitational force, the correct compression of the spring was determined to be 0.8 meters. This highlights the importance of careful attention to all variables in energy conservation equations. The user expressed relief upon realizing the mistake and successfully finding the correct solution.
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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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