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Chapter 7 Post Homework

# Chapter 7 Post Homework - Since all energy of the sled is...

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Chapter 7 Post Homework: #2 For this type of question you need to look at the energy of the moving object. If we set the lowest point (B and D) to have 0J Potential energy then the highest point (where the sled starts) has a potential energy of mgh = (210)(9.8)(50) = 102900J. Since friction is zero all the way to point D, the energy of the sled and its passengers is maintained up to that point. At point D we can see that it has 0J of potential energy, so lost 102900J. In order for it to have the same energy it must therefore have gained 102900J Kinetic energy. Starting at point D, the sled gradually loses its kinetic energy till it comes to rest The energy lost is the work done by friction = Fd The friction is equal to the product of the normal force (mg) and the coeff. of friction. F = 0.4(210)(9.8) = 823.2N

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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 = M1a-M2a + 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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