Physics Help Forum Spring Compression
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 May 2nd 2009, 06:26 PM #1 Junior Member   Join Date: Apr 2009 Posts: 25 Spring Compression Question: A 1.00 kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P7.54). The object has a speed of vi = 3.00 m/s when it makes contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring's unstretched position. The object finally comes to rest a distance D to the left of the unstretched spring. Find the distance of compression d. Alrighty, so I did the normal energy "recipe" of 1/2mv^2 - (mu)mg(x) = 1/2kd^2 However, no matter which way I try to solve it I run into the problem that I have two distance variable: x (from the initial position of the block to the block against the compressed spring) and d (the distance the spring is compressed). I'm not sure how to correlate these together and no inital distance is provided. Any ideas? Thanks a bunch!!
 May 2nd 2009, 07:36 PM #2 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 You need not worry about the initial position of the block. Start from the situation given in the second figure. The velocity 3.0 m/s is just before the object makes contact with the spring. So (1/2)mv^2 is the kinetic energy just before the contact. Now if the spring is compressed by a distance d then work done by friction will be (mu)mg(d) and not (mu)mg(x). Except this extra bit of salt your recipe is fine. So- (1/2)mv^2 - (mu)mg(d) = (1/2)kd^2 For the final part you should have- (1/2)kd^2 - (mu)mg(D+d) = (1/2)kD^2
 May 3rd 2009, 12:36 PM #3 Junior Member   Join Date: Apr 2009 Posts: 25 Thank you Parvez!

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