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Unformatted text preview: Since all energy of the sled is lost when it comes to rest we have: Work done by friction = 102900N = Fd> 823.2d = 102900> d = 125m #3 T  M2g = M2a (taking up as positive and down as negative). .. So T  M2g = M2a T = M1aM2a + M2g = M1a M1a + M2a = M2g a(M1+M2) = M2g a= M2g / (M1+M2) a=3.27. .. so T = M1*a = 6*3.27= 19.6N So work = 19.6N*1m = 19.6J. .. hope it's right. #6 Well, assuming that only the mass changes and everything else stays constant. .. Ek = 1/2mv² Therefore doubling the mass doubles the kinetic energy, and we have twice as much energy to dissipate. However, the frictional force is proportional to the normal force, which in turn is proportional to mass. .. thus you're going to dissipate energy twice as fast while braking. .. so I believe the answer is the same distance....
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 Spring '09
 Heins
 Energy, Force, Kinetic Energy, Mass, Potential Energy

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