# Homework Help: Simple mechanical energy question [gr.11]

1. Mar 22, 2004

### kassy_in_the_sky

[urgent] simple mechanical energy question [gr.11]

Several children, pretending they are playing in the jungle, suspend a rope from an overhead tree limb. A child of mass 40 kg running at 8.0m/s grabs the rope and swings off the level ground.

a) What maximum height does the child reach?
b) How fast would a 30 kg child have to run to reach the same height as the 40 kg child?

½mv1²+mgh1 = ½mv2²+mgh2

a) m=40kg, v1=8.0m/s, h1=0m, g=9.8m/s², v2=0m/s (at max height v=0)

h2=?

½mv1²+mgh1 = ½mv2²+mgh2

½mv1²= mgh2

h2 = 3.27m = 3m (is this correct?)

im really confused with B... coz i got 8m/s... it doesn't make sense to me how they can run at the same speed and achieve the same height??!!

Last edited: Mar 22, 2004
2. Mar 23, 2004

### dorthod

A)
I would say: the KE at the point of him grabbing it = the PE at his peak(all his KE has become PE) so KE=PE so ½mv²=mgh substituting:
.5*40*8²=40*9.8*h
1280=392h
h=3.265m

B)change 30 for 40 above and v is unknown
.5*30*v²=30*9.8*3.265
15v²=959.91
v²=63.994
v=7.999m/s

yes, it is the SAME v for both! Why? because the m cancels out on both sides of the equation: the m in ½mv² and the m in mgh divide out!

Last edited: Mar 23, 2004
3. Mar 23, 2004

### kassy_in_the_sky

thank you!! ^^