100 kg elevator acclerates up at rate avg a*. What's the avg force acting on the elevator if it covers a dista?

Question

100 kg elevator acclerates up at rate avg a*. What's the avg force acting on the elevator if it covers a distance X over a period of 10s?

a* denotes average accleration
a. 2X + 1000
b. 100(a* + 1)
c. 2a* + 1000
d. 2a*X

I set up a free body diagram to show all the forces acting on the elevator. There's a downward force due to gravity, Fg, the net upward force, Fn, causing the accleration.

Fn - Fg = ma*
Fn = ma* + Fg

The question states that it travels a distance of X over 10 sec

a* = delta V/delta t
delta V = X/t , which the question gives me
delta V = x/10s, i substituted this into the a* equation and got
a* = x/100s

when i plug this into my net force equation:

Fn = ma* + Fg
Fn = m(a* + g)
Fn = 100kg(x/100s + 10m/s^2)
Fn = X + 1000

I can't seem to figure out why my answer doesn't match any of the answer choices. Can someone please point out what i'm doing wrong?
Thanks

please show work.

Answer 1

They didn't tell you how acceleration was not constant, all they told you is that it isn't necessarily constant.



Because you don't really have enough information to produce the comprehensive kinematics graphs, they are loosely using the term "average acceleration" as if it were the acceleration if acceleration somehow were constant.



Also, "average force" is a term which means whatever force is required to give this "average acceleration".



This is bad physics practice to do so, and can lead to great confusion, so you really should call shenanigans on your instructor.



A problem should state "assume acceleration is constant" if they want to keep it simple. AND if acceleration is not constant, enough information should be given to state otherwise.









Another thing, they are just asking for "the avg force acting on the elevator".



That is so VAGUE!!!!!!



What the hell force are they talking about? They just specified force acting on the elevator.



There are two forces acting on the elevator. Tension in the lift cable AND Earth's gravity.



WHICH of those is of interest? OR is the NET FORCE of interest?





They didn't specify. Call shenanigans like you've never done before.

Answer 2

You wrote:



> "delta V = X/t , which the question gives me"



This is incorrect. "X/t" equals the AVERAGE velocity; this in general is quite different from the CHANGE in velocity.



And in any case, I think this is irrelevant to the problem. The only sensible definition of "average force" that I can think of is: "mass times average acceleration," i.e. just Newton's 2nd Law. In that case,



F* = m(a*) = 100(a*)



Of course, that's not one of the answer choices. :-(. So maybe what they really meant was: "Average UPWARD force". In that case we have:



F_net_avg = F_upward_avg - Weight = m(a*)



Or:



F_upward_avg = m(a*) + Weight

= m(a*) + mg

= m(a* + g)

= 100(a* + g)



That looks KIND of like answer choice "b", but not quite.



Okay, let's try one more thing, and pretend the acceleration is constant. In that case we can say:



X = (v1)t + ½(a*)t²



And (since they didn't specify v1) if we further pretend that v1 is zero, we can say,



X = ½(a*)t²



Whence,



a* = 2X/t²



and



F* = m(2X/t²)



If you plug in the numbers "100" for m and "10" for t (and ignore the units), you get:



F* = 2X



Still not one of the answer choices. I agree with gintable that this is a hopeless, poorly constructed problem.

Answer 3

The last person is correct. The answer is Fnet = 2x. The Kaplan MCAT book is wrong, as are the first few "answers" here. F = 2x + 1000 is the tension force in the supporting cable. GA!

Answer 4

hhahhahaha...my kinfolk in the subsequent door are questioning why i fell from the chair and crying and tearing in the comparable time.. thx for the snort... thats the main suitable one *carry the doorways open and say you're arranged on your chum. After awhile, permit the doorways close and say, "hi Greg. How's your day been?"*