# Homework Help: Work / Energy Elevator / Spring Problem

1. Nov 23, 2016

### tristanslater

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Not 100% sure, but probably: 1/2mv^2, PE = mgh, -1/2kx^2

3. The attempt at a solution
I am pretty lost on this one, but have been attempting to figure out the spring constant based on the information that the elevator is stopped at 2 m. I tried F_net = F_spring - F_g + F_f, F_net at 2 m is 0, so F_spring = F_g - F_f, F_f is given, and F_g is easy to workout (mg), from there I get: k = (mg-F_f) / delta-y, which gets me 2600 as a spring constant. I also calculated the KE_i = 1/2m(v_i)^2.

After that I get kind of stuck. I feel like I am supposed to figure out the change in energy, but I'm not sure how friction and the spring factor into that.

Thanks,
Tristan

2. Nov 23, 2016

### oz93666

Energy balance ... you know the kinetic energy at first contact.... energy lost in friction brake is force x 2m..... potential energy from 2m. drop... from these knowns you can find energy stored in spring and force in spring at all points (linear) ...

Now you have enough to find speed at 1m.

3. Nov 23, 2016

### Staff: Mentor

Hi tristanslater. An energy approach is the way to go.

You won't get the spring constant by looking at the forces since the elevator is not in equilibrium when the spring is fully compressed. In fact, the elevator will be poised to "bounce" at that instant, being accelerated upwards by the spring force.

For these sorts of problems it can help to start out with a list of all the places that energy can come from or go to in the given scenario. Here you've got kinetic energy, gravitational potential energy, spring potential energy, and frictional loss You've listed relevant equations for each of these except for the friction energy loss (work done by friction), although your spring potential energy expression should not have the minus sign.

Then it's time to write the energy balance equation for the two instants of interest (first contact with spring, and zero velocity instant with spring compressed 2 m). Give it a try.

4. Nov 23, 2016

### tristanslater

K_f + U_f = K_i + U_i
Oh! You're right! I was thinking no motion = equilibrium, but you're right.
I'm thinking the equation should be PE_f + KE_f = PE_i + KE_i - W_ext. I think this makes sense, because if there was no external forces, they would be equal, U lost would become K as the elevator sped up, but now W is being done against that falling. At the farthest compressed, PE_f = 0 (have PE relative to fully compressed) and KE_f = 0 (not moving), so this gives us W = PE_i + KE_i, which makes sense, enough work would need to be done to cancel the energy it already had. The firction would also be 0, because friction opposes motion and there is no motion. That means the only work would be: W = 1/2k*y^2. Since, y is 2, W = 2k = 1/2m(v_i)^2 + mgh_i => k = 1/4m(v_i)^2 + mg, which I work out to be k = 27600. Does this look right?

Now for when it is compressed by 1 m. Start the same PE_f + KE_f = PE_i + KE_i - 1/2ky^2 - (F_f)y => mgh_f + 1/2m(v_f)^2 = mgh_i + 1/2m(v_i)^2 - (F_f)y - 1/2ky^2

I figured then you solve for v_f, but when I do, the numbers don't come out right.

Am I on the right track?

Thank you,
Tristan

5. Nov 23, 2016

### Staff: Mentor

I'm not fully following your argument, particularly when you write: "W = 2k = 1/2m(v_i)^2 + mgh_i ..." If k is the spring constant then 2k would not be work, even if the 2 is a distance in meters. That would be Hooke's law giving force, not Joules. And I'm not getting your value for the spring constant.

The way I approach these problems is to set on one side of the equation all the places where energy comes from between the initial state and final state, and on the other side everywhere it goes or ends up. So between the instant when the elevator first hits the spring and when it ends up at its lowest point it starts with the initial KE and gains more energy via the change in gravitational PE as it falls. Then on the other side of the equation the energy ends up as spring PE and friction loss (no residual KE in this case since we're looking at the instant when the elevator's downward motion has just halted). So the big picture is:

$KE_i + ΔPE_g = ΔPE_s + W_f$ ,

where the g and s subscripts indicate gravitation and spring, and $W_f$ is the work done by friction. Each of the terms can then be expressed using the appropriate relevant equation. The spring constant gets introduced when the $ΔPE_s$ gets replaced with its corresponding formula.

When the "final" position of interest is one where there is still motion going on (like the 1.0 meter point), then there will be a $KE_f$ term on the right hand side indicating that some of the energy ends up as (or remains as) KE.

6. Nov 23, 2016

### tristanslater

I didn't realize we could use latex in these forums. That helps a lot!

I think we're actually writing the same thing, just slightly differently.

When I wrote: $W = 2k = \frac{1}{2}mv_i^2 + mgh_i$, I was just shortcutting. Spring potential energy: $\frac{1}{2}k\Delta y^2$, when $\Delta y = 2 \Rightarrow \frac{1}{2}k(2)^2 = \frac{1}{2}k(4) = 2k$ The spring potential should be equal to the total energy at the start, so $2k = \frac{1}{2}mv_i^2 + mgh_i$, solving for $k$, we get: $k = \frac{1}{4}mv_i^2 + \frac{1}{2}mgh_i$

I figured it out. Where I was going wrong was that I was assuming that since there was no motion there is no friction force at the bottom, but we are talking about work done from $h = 2 \Rightarrow h = 0$, so work done by friction is $F_f\Delta y$, since $\Delta y = 2 \Rightarrow W_f = 2F_f$. Plugging that back in for $k$, we get: $k=\frac{1}{4}mv_i^2 + \frac{1}{2}mgh_i-F_f$. When I use this value for $k$, everything else works out.

I think your formula is actually the same as mine. We're just approaching it differently.

Thank you,
Tristan