Ore cars with total mass M = 1.2*10^4 kg start from rest and roll down the track

Question

Ore cars with total mass M = 1.2*10^4 kg start from rest and roll down the track to an unloading area 10.0 m below, as shown in Figure 8.35. They are stopped by a spring bumper and held with a clamp while they are unloaded.

I have a image of the Figure. Sorry its Blurry:
//tinypic.com/r/35iwt8x/7

1)What spring constant is needed if the bumper compresses by 1.3 m when stopping the ore car?.

2)After a car is unloaded, its remaining mass is 1.9*10^3 kg. If after the clamp was released the empty car was allowed to return back up the track, with what speed would the car reach the top?

Explanation / Answer

mass of the ore car is M = 1.2*104 kg The height of release is h = 10 m a) The compression in the spring is x = 1.3 m Let k be the spring constant The potential energy of the spring is U = 0.5kx2 As the car is at rest initially, it has only potential energy given by U = mgh According to the law of conservation of energy mgh = 0.5kx2 k = 2mgh/x2 k = 2(1.2*104 kg)(9.8 m/s2)(10 m)/(1.3 m)2 k = 1.39*106 N/m b) The potential energy stored in the spring is U = 0.5kx2 U = 0.5(1.39*106 N/m)(1.3 m)2 U = 1.174*106 J With this energy, the empty car will have kinetic energy, K = 1.174*106 J (conservation of energy) Let v be the speed of the car at the top Then the conservation of energy yeilds mgh + 0.5mv2 = K (1.9*103 kg)(9.8 m/s2)(10 m) + (0.5)(1.9*103 kg)v2 = 1.174*106 J v = 32.246 m/s U = 0.5(1.39*106 N/m)(1.3 m)2 U = 1.174*106 J With this energy, the empty car will have kinetic energy, K = 1.174*106 J (conservation of energy) Let v be the speed of the car at the top Then the conservation of energy yeilds mgh + 0.5mv2 = K (1.9*103 kg)(9.8 m/s2)(10 m) + (0.5)(1.9*103 kg)v2 = 1.174*106 J v = 32.246 m/s

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