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Introductory Physics Homework Help
Motion with gravitational, elastic, and friction forces
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[QUOTE="BvU, post: 4988446, member: 499340"] 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. [B]free fall[/B] 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 [U]magnitude[/U] of ##\vec g\;,\quad## which is a positive value ( ##|\vec g| = ## 9.8 m/s[SUP]2[/SUP] ). In words: if it drops (meaning: if h decreases, so ##\Delta##h is negative), kinetic energy [U]increases[/U]. This increase of kinetic energy is delivered by the gravitational force: gravity [U]does positive work[/U]. 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!) ) [B]braking[/B] 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 [U]negative[/U] 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 [U]by[/U] the elevator, and (I hope, but I'm pretty sure) W[SUB]other[/SUB] in henry's expression means the work done [SUB]on[/SUB] the elevator, other than from gravity. [B]spring[/B] 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 [U]on[/U] the elevator, elevator does positve work [U]on[/U] the spring, so energy is transferred from elevator to spring. And the equation grows to $$ E_{kin} + E_{pot} + {Work}_{\rm on\ brakes} + {Work}_{\rm on\ spring}= {\rm constant} $$ [B]combination[/B] 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): [INDENT]K[SUB]i[/SUB] + U[SUB]i[/SUB] = ##{1\over 2} mv^2+ mgh = {1\over 2} mv^2##[/INDENT] At h = -2 m when the thing comes to a standstill [INDENT]K[SUB]f[/SUB] + U[SUB]f[/SUB] + Work[SUB]done by elevator[/SUB] = ##\ \\ \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## [/INDENT] Signs still look a bit awkward, so that goes to endorse Simon's good advice ! [B]epilogue[/B] 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) [/QUOTE]
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Motion with gravitational, elastic, and friction forces
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