# Motion with gravitational, elastic, and friction forces

## Homework Statement

A 2000-kg elevator with broken cables in a test rig is falling at when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17,000-N frictional force to the elevator. What is the necessary force constant k for the spring?

## Homework Equations

Wother + Ki+Ui = Kf+Uf

## The Attempt at a Solution

Wother + Ki = Uf
(-17000 * 2) + (1/2)(2000)(42) = (1/2)k(22) + (2000)(-9.8)(-2)

So the correct answer to the problem is k = 10600 N/m. My book solves this by making the final potential energy negative so that it is (2000)(9.8)(-2) and then solving for the spring constant. What I don't understand is why the force of gravity is not negative to make the entire final potential energy positive. The negative 2 clearly states that down is negative so why is gravity not negative if it is also pointing down?

Simon Bridge
Homework Helper
Energy does not have a direction. Gravity may point down but is gravitational potential energy lost or gained and by what?

You can get yourself tied up in knots thinking about how to fit the problems into the equations you have. The way to think about these problems is in terms of the energy transformations involved -

i.e. What sort of energy does the lift lose, and where does it go?

- that way you get to fit the equation to the problem, which is actually best practice.

BvU
Homework Helper
Please fix the "falling at when" in the problem formulation... Is it really 4 m/s ? So it's been dropped from 0.8 m and it compresses the spring by 2 m ? I thought buildings were a little higher :) (but I agree that that's the problem maker's responsibility, not the problem solver's)

If you lift the 2000 kg by 2 m, do you increase its potential energy or do you decrease it ? So does gravity do positive or negative work ?
If you lower the 2000 kg by 2 m, do you increase its potential energy or do you decrease it ? So does gravity do positive or negative work ?

In fact, potential energy is ##\ \ \displaystyle -\int \vec F \cdot \vec {dx} = -mg\Delta h## and in that last expression g is -9.8. Sloppy popular language is then U = mgh.

The work gravity does has to be dissipated/absorbed by clamp/spring.

If down is negative, you want to write 17000 * -2 in order not to confuse the reader...

haruspex
Homework Helper
Gold Member
2020 Award
Is it really 4 m/s ?
Clearly not, since it makes the spring redundant.

BvU
Homework Helper
Clearly not, since it makes the spring redundant.
Beg to differ. Clamp alone (17000 N) can't offset gravity force (19600 N). But I was wrong to assume that the clamps only start acting when the elevator is at the top of the spring. Still, the 4 m/s is then reached after only 3 seconds when the acceleration is (196000 - 17000)/2000 = 1.3 m/s and that would be after some 6 m. Not much.

Simon has good pragmatic advice, but it would be nice if we (including henry! -- especially henry, in view of his (/her?) preceding posts) could sort this out in a way that it's also OK formally.

free fall

In free fall, the elevator acquires kinetic energy which is provided by gravity. The balance looks like $$E_{total} = {\rm constant}\qquad E_{total} = E_{kin} + E_{pot} = {1\over 2} mv^2 + mgh$$And we fill in the magnitude of ##\vec g\;,\quad## which is a positive value ( ##|\vec g| = ## 9.8 m/s2 ).
In words: if it drops (meaning: if h decreases, so ##\Delta##h is negative), kinetic energy increases.

This increase of kinetic energy is delivered by the gravitational force: gravity does positive work. In formula: ##\vec F## and ##\vec ds## are in the same direction, so their vector inner product is positive. The elevator "does negative work" (it is being accelerated by a pulling force - it exercises a fictitional inertial force in the direction opposite to the acceleration. Believe it or not: it is pulling the earth towards the elevator with a force mg (19600 N upwards!) )

braking

The clamps act as brakes. They exercise an upwards force and as long as there is movement they convert kinetic energy into heat. I.e. they do negative work: force and motion are in opposite direction. The energy is taken away from the kinetic energy of the elevator, which therefore does positive work (it pushes down while moving down). Equationwise: $$E_{kin} + E_{pot} + {Work} = {\rm constant}$$Looks different from henry's equation but it is not. For the simple reason that my Work is the work done by the elevator, and (I hope, but I'm pretty sure) Wother in henry's expression means the work done on the elevator, other than from gravity.

spring
The spring acts as the brakes, only it doesn't convert into heat but into mechanical energy in the compressed spring. So the same signs apply: spring does negative work on the elevator, elevator does positve work on the spring, so energy is transferred from elevator to spring. And the equation grows to $$E_{kin} + E_{pot} + {Work}_{\rm on\ brakes} + {Work}_{\rm on\ spring}= {\rm constant}$$

combination
Putting everything together we have

At h = 0 m, the top of the spring (an arbitrary choice, we have the freedom to pick a zero point):
Ki + Ui = ##{1\over 2} mv^2+ mgh = {1\over 2} mv^2##​

At h = -2 m when the thing comes to a standstill
Kf + Uf + Workdone by elevator = ##\ \\ \qquad
{1\over 2} mv^2+ mgh + \vec F_{\rm against\ friction} \cdot \vec {\Delta \rm h} + {1\over 2} k\; {\Delta \rm h}^2 = \\ \qquad
m (9.8 {\rm m/s^2}) * (-2 { \rm m}) + 17000 * 2 \;{\rm m} + {1\over 2} k\;(2\; {\rm m})^2##
Signs still look a bit awkward, so that goes to endorse Simon's good advice !

epilogue

At the end of this dramatic story, the elevator is hopefully not moving any more, with the spring exercising 21200 N upwards and gravity 19600 N downwards, The clamps now have to deliver 600 N downwards, so no longer the 'constant' 17000 N
(either these 600 N down, or the elevator will have to move up a little)

As a physicist with near-zero knowledge about elevators, I would put in a plea to to beef up the braking force of the clamps, so that (if the cable breaks at the tenth floor)