Ch6 problem45

Physics: Principles with Applications

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Ch6 problem45 Solution: The total energy of the elevator and spring is conserved. Initially, the energy of the spring is 0 J, since it is uncompressed. Assuming that the elevator is at rest when the cable breaks, the initial kinetic energy of the elevator is also 0 J. The initial potential energy of the elevator, however, is no zero, but PE e = Mgh. So, the total initial energy TE 0 = Mgh. Now, just before the elevator hits the spring, the total energy of the spring is still 0 J. Since the elevator is now fallen to a height of 0 m, the potential energy of the elevator is now also 0 J. However, it does have a kinetic energy of KE e = 1=2Mv 2 . So the total energy at that moment is TE 1 = 1=2mv 2 . By the conservation of energy, these must be the same, so Mgh = 1 2Mv 2 ; or, solving for v, v = p 2gh: Now, when the elevator hits the spring, it compresses it by an amount x before coming to rest. This causes an acceleration a which is related to the distance of the compression
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Unformatted text preview: (the distance over which the acceleration occurs) by 0m=s 2 = v 2 + 2ax; or, substituting in the expression we got for v above and solving for x, x = h g a h: If we want the acceleration a to be at most 5g, then we need to have x = h h=5. At the same time, we must have conservation of energy again. After the elevator has come to rest again and the spring is compressed, the kinetic energy of the elevator is 0 J. The elevators potential energy is now PE e = Mgx. The potential energy of the spring is PE s = 1=2 kx 2 . The total energy is therefore TE 2 = Mgx + 1=2 kx 2 . By the conservation of energy, this must be equal to TE , the energy just before the elevator starts compressing the spring. So Mgh = Mgx + 1 2kx 2 : Substituting in x = h=5, as we found just above, makes this equation Mgh = Mg(h h=5) + 1 2 k (h h=5) 2 : Solving this for k, we get k = 60 Mg h :...
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