# Conservation of Energy Problem-I think

bbbbbev
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

## Answers and Replies

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?

bbbbbev
Thanks! I had forgotten to multiply by g somehow!! I can't believe I didn't notice that before!