Physics (Grade 12)?

Question

An Elavator of Mass 800kg is descending at a speed of 4.0 metres per second. Find the shortest distance over which this elavator could be stopped if the greatest force the cable can withstand is 11000 newtons?

What Is the Shortest Distance It Can Be Stopped?

Answer 1

force to maintain speed is 800*9.8=7840N so only 11000-7840=3160N available.
F=Ma or
a=F/M=3160/800=3.95m/s^2
v=0=4-3.95t
t=4/3.95=1.01s
d=v0t-.5at^2=4-2=2 m

Answer 2

Lets look at the elevator at rest.

f = ma
f = 800 times gravity, assume it is 10 because i dont have a calculator.
f = 8000 newtons.

For the elevator to at least stay put, the tension of the cable is also 8000 newtons in the opposite direction, namely up.

A decending elevator means the gravitational force is greater than the tension of the cable. This means, at any time the elevator is descending, it will not exceed 8000N tension on the cable, much less than 11,000.

Only an ascending elevator would have a tension exceeding the gravitational force and come close to 11000 newtons.

Answer 3

bit small solution

f = m *a
max acc. acting on lift is ~10 m per second square
hence max force is 10*8000 = 8000 N which can never exceed 11000 N

when velocity is constant no force is applied, know why
change in momentum w.r.t time is zero

Answer 4

f=ma a=f/m=11000/800=110/8. d=u^2/2a from the third eqn of kinematics. =4^2/2(110/8)

Answer 5

i think you can use these equations

f=ma, v^2=v0^2+2*a*x

crap i wrote the equations wrong, the third guy got it right I fixed it though.

Answer 6

http://www.physics.uoguelph.ca/tutorials/tutorials.html
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html