In a \"worst-case\" design scenario, a 2000-kg elevator with broken cables is fa
ID: 1427328 • Letter: I
Question
In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator. What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring? When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?Explanation / Answer
Lets start with the force of gravity on the elevator.
2000 * 9.8 = 19600N
But the friction clamp opposes this with a force of 17000 N
So the Net force on the elevator is 19600 - 17000 = 2600 N
At the point where the elevator contacts the spring it will have an energy 1/2 m v^2
= 1000 * 4^2 = 16000J
During the next two metres the spring will a) absorb the kinetic energy and b) absorb the gain in energy due to the net force.
= 16000 + F*d
= 16000 + 2600 *2
= 21200 J
Now to get the spring constant.
1/2 k x^2 = E = 21200
k = 21200 * 2 / x^2
= 21200 * 2/4
= 10600 N/m
To find the acceleration at 1 m compression, the force provided by the spring = Kx
= 10600 * 1 = 10600 N ( upward)
Add this to the other two forces ( mg and friction which together give 2600 down)
and the resultant force is 8000 N upwards.
F = ma
so a = f/m = 8000 / 2000 = 4m/s^2 upwards at this point.
Now given that my spring constant does not agree with yours I doubt that I would get 3.65 m/s at the compression of 1m
original Kinetic energy + work done = final kinetic energy + spring energy
16000 + 2600 = 1/2 m v^2 + 1/2 k x^2
18600 = 1/2 m v^2 + 5300
1/2 m v^2 = 13300
v^2 = 13.3
v = 3.65 m/s