Question

An elevator cable breaks when a 970kg elevator is 23m above a huge spring (k = 2.1×105 N/m ) at the bottom of the shaft. Part A)Calculate the work done by gravity on the elevator before it hits the spring. B)Calculate the speed of the elevator just before striking the spring. c)Calculate the amount the spring compresses (note that work is done by both the spring and gravity in this part). We only went through part of our work chapter and this is the very last question and I do not know how I would set these steps up. Any help is appreciated, please and thank you!

Answer 1

I am not master in these type of questions...
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do study this

Answer 2

I'm not 100% sure about this but here's a shot :)

a) Work=Force times Distance, so it follows that work done on elevator=970kg*9.81m/s^2*23m=218861.1 joules

b) Take the energy from a) and equate it with kinetic energy: 218861.1J=.5mv^2 ; where m=970kg. Solving for v, you get approximately 21.24m/s.

c) Your potential energy for the spring = kinetic energy of elevator before hitting the spring + work done one elevator due to gravity. So you have: .5kx^2=218861.1J + mgx; where k is your spring constant, m is the mass of elevator, g is acceleration due to gravity, and x is the distance compressed. Solving for x (distance compressed), you get x=1.5m

Answer 3

This is a conservation of energy problem. Let's make a couple of assumptions to make the problem more manageable.

1) No air resistance.
2) No friction.
3) The elevator is at rest when the cable breaks.


For part a, let's recall the definition of work due to a constant force (which gravity is). \[W = F \cdot d\]where F is the force and d is the distance over which the force acts. Here, \(F=m g\)

Part b. Since energy is conserved here, we can say that\[\Delta PE = \Delta KE \rightarrow mgh = {1 \over 2} mv^2 \rightarrow gh = {1 \over 2}v^2\]

Part c. Let's recall that the work of a spring is given as\[W = \int\limits F dx = \int\limits kx dx = {1 \over 2} kx^2\]Things get a bit tricky here conceptually. The problem statement suggest that we should simply say \[W_s = W_g\]which becomes\[kx^2 = mv^2\] But gravity continues to do work on the elevator while compressing the spring. A more robust solution would be to consider the Work-Energy theorem\[W = \Delta PE+ \Delta KE\]Then we have\[{1 \over 2} k x^2 = {1 \over 2} mv^2 + mgx\]

Take your pick here.