Two objects with masses 4.90 and 2.10 hang 0.800 above the floor from the ends of a cord 7.20 long passing over a frictionless pulley. Both objects start from rest.

Question

Find the maximum height reached by the 2.10 object.

The heavier object will move .8 m down, the lighter will move up an additional .8m.

max height= initial height+movement up.

but how do u find the initial height?

Thanks

Answer 1

Hey Ben,
alright first you have to determine the acceleration of the system.
Let m1=2.1kg and m2=4.9kg
Use
F=ma
m2g-m1g=m2a+m1a
(m2-m1)g=(m2+m1)a
a= (m2-m1)(g)/(m2+m1)
= (4.9-2.1)(9.8)/ (4.9+2.1)
=3.92 ms^-2
Now we must determine the speed of m1
we use
v^2=u^2+2as
we know that m1 was at rest so u=0 m/s
and also we know that m2 will go all the way down so m1 moves up 0.8m so s=0.8 m
v^2=0+(2)(3.92)(0.8)
v^2= 6.272
v=2.5044 m/s

Now we know that m1 is (0.8+0.8)m above the ground as m2 goes down to the floor

So we add 1.6m to the max height

To calculate max height
use formula
h=V^2/2g
= (2.5044)^2/ (2 x 9.8)
=0.32m
so add 1.6m to max height
0.32+1.60=1.92m

HOPE THIS HELPS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

"so add 1.6m to max height
0.32+1.60=1.92m"

Sorry I meant max height is = 0.32+1.60=1.92m