Work done on a falling elevator by a spring + gravity

• simphys
In summary, the problem is that the second part of the motion has an incorrect initial velocity, and the final velocity should be zero.
simphys
Homework Statement
An elevator cable breaks when a 925-kg elevator
is 22.5 m above the top of a huge spring
at the bottom of the shaft. Calculate
(a) the work done by gravity on the elevator before it hits the
spring; (b) the speed of the elevator just before striking the
spring; (c) the amount the spring compresses (note that here
work is done by both the spring and gravity).
Relevant Equations
WE principle + w=Fd
I don't understand what I have done wrong in part (c) I have the initial velocity for the second part of the motion and have the final velocity zero and then the net work done is W_mg + W_Fs and the actual answer for x is 2.37m

Here is my solution

small edit: After getting V_2 = 21m/s, x = 3.19m and the ,(comma) is a decimal point here! .(dot) is multiplication

Last edited:
Delta2
I think I see the problem.. I used the work done by gravity from the free fall motion part.
for the second part it actually is W_mg = mgx (if I am not mistaking)

I got the following: ##-4E4*x^2 + 9.074E3*x + 2.04E5=0 ##
which I solved with the calculator directly, but using the quadractic formula would be a lot of writing, is there another way I can look at this problem or something?

simphys said:
I got the following: ##-4E4*x^2 + 9.074E3*x + 2.04E5=0 ##
which I solved with the calculator directly, but using the quadractic formula would be a lot of writing, is there another way I can look at this problem or something?
I cannot read the value of k in the posted image, but I deduce it must be 8E4 N/m.

simphys
haruspex said:
I cannot read the value of k in the posted image, but I deduce it must be 8E4 N/m.
Thank you! I thought that there may be a way to just use the initial and final position instead of breaking it up into part, but I didn't try. I supposed that it might've work that is why I asked.

simphys said:
Thank you! I thought that there may be a way to just use the initial and final position instead of breaking it up into part, but I didn't try. I supposed that it might've work that is why I asked.
Yes, if you only had to solve part c you could just write the conservation of energy equation straight off:
##(h+x)mg=\frac 12kx^2##
and get the quadratic from that.

simphys
haruspex said:
Yes, if you only had to solve part c you could just write the conservation of energy equation straight off:
##(h+x)mg=\frac 12kx^2##
and get the quadratic from that.
Oh right, way faster thus. Thank you!

1. How does a spring and gravity affect the work done on a falling elevator?

The spring and gravity both contribute to the work done on a falling elevator. The spring exerts a force on the elevator, causing it to compress and store potential energy. Gravity also exerts a force on the elevator, causing it to accelerate downwards and convert potential energy into kinetic energy. Both of these factors contribute to the overall work done on the elevator.

2. What is the relationship between the work done and the distance traveled by the elevator?

The work done on a falling elevator is directly proportional to the distance traveled by the elevator. This means that as the elevator falls, the work done by the spring and gravity will increase in proportion to the distance it falls. This relationship can be described by the equation W = Fd, where W is work, F is force, and d is distance.

3. How does the mass of the elevator affect the work done by the spring and gravity?

The mass of the elevator does not directly affect the work done by the spring and gravity. However, it does affect the amount of potential and kinetic energy that the elevator has, which in turn determines the amount of work done. A heavier elevator will have more potential energy and require more work to be done by the spring and gravity to reach the same distance as a lighter elevator.

4. Can the work done on a falling elevator be negative?

Yes, the work done on a falling elevator can be negative. This would occur if the elevator is slowing down or moving in the opposite direction of the force exerted by the spring and gravity. In this case, the potential energy of the elevator is decreasing and the work done by the spring and gravity is negative.

5. How does the speed of the elevator affect the work done by the spring and gravity?

The speed of the elevator does not directly affect the work done by the spring and gravity. However, it does affect the amount of kinetic energy that the elevator has, which in turn determines the amount of work done. A faster elevator will have more kinetic energy and require more work to be done by the spring and gravity to reach the same distance as a slower elevator.

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