**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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- Energy, Potential Energy, 0m, 0 m, 0 j, 1 2 k