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Gagan_Bhullar@dwc.com (Gagan Bhullar) wrote:

>1.	A seventeen meter long vine hangs vertically from a tree on one side of a
>ten meter wide 						gorge.  Tarzan runs up hoping to grab the vine, swing
>over the gorge and drop vertically off the vine to land on the other side. 
>At what minimum speed must he be running?

>2.	At the bottom of a frictionless ski slope is a 20 meter wide stretch of
>rough snow where the coefficient of friction is 0.32.   From what vertical
>height on the slope should a skier start at rest in order to get through the
>rough patch with a speed of 6.0 m/s.

>It would be much appreciated if I could get some form of input by the 14 of
>October!
>Thanks in advance
Dear Gagan

These questions are both energy conservation questions.  They require
that you know that an object converts potential energy  (mgh) to
kinetic energy (1/2 m v^2) as it falls.

1.  In the first problem the end of the vine will rise to a height h
as it moves over the gorge.  This is why Tarzan has to drop off.

To get the height  you need the theorem of Pythagorus and a simple
diagram.  This gives:  (17 - h)^2 + 10^2 = 17^2
From this (and the quadratic formula) h = 3.252273 metres
Using energy conservation:  1/2 m v^2 = mgh
thus 1/2 v^2 = (9.8)(3.252273)
and v = 7.9840 m/s   or, rounding, v = 8.0 m/s

2.  Here initial energy (mgh) = energy lost to friction + final
kinetic energy.

    mgh = .32 (mg)(20)  + 1/2 mv^2
    9.8 h = .32(9.8)(20) + 1/2(6)^2
     h = 8.2367346 = approx. 8.2 metres

Note that the force of friction = (coefficient)(normal force) 
In this case coefficient = .32 and normal force = weight = mg

Sorry I am late in replying but I thought I would post anyway.  I will
try to be quicker next time and hope you keed the questions coming.

David Stock