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December 18, 2014

December 18, 2014

Posted by **Tom** on Tuesday, January 31, 2012 at 12:45pm.

1. Jenny (75 kg) goes on a ski trip. She goes down a slope that's inclined at 15*. Her drag coefficient is 0.300, the area of contact with the air is 0.750 m^2 and the air density is 1,31 kg/m^3. The coefficient of kinetic friction between the skis and the snow is of 0,185.

Jennys speed according to her distance:

from 0 to 15m/s: between 0m and 50m

from 15 to 20 m/s: between 50m and 100m

constant 20m/s: 100m to 450m

from 20m/s to 0: 450m to 500m

Questions:

a) What is the Work done by the force of air resistance?

A) I answered 0.687 J which seems very unlikely. This is how I proceeded:

I have the Force of air resistance formula:

1/2 CpAv^2

I have all of the numbers except for the velocity. So I just used the graph to calculate the total time it took Jenny to complete the slope. It gave me: 37.83 seconds. I then proceeded to divide the total length of the slope which is 500m. by 37.83s. to get her average velocity for that descent. It gave me 13.28 m/s which I then used in the air resistance formula and got 26N of air resistance against her. Then I just went with

W = J/s ---> W = 26 N / 37.83 s.

and it gave me 0.687 J.

Could this possibly be the answer or did I do something wrong?

- Physics -
**Tom**, Tuesday, January 31, 2012 at 6:00pmso I was supposed to do Fres. X distance

for 1 J = 1N*m. I had mixed the previous formulas up, I had the W for watts instead of Work. So my final answer for the work of the air resistance is actually 26N*500m = 13,000 J or 13 kJ.

is that correct?

- Physics -
**Tom**, Tuesday, January 31, 2012 at 6:02pmFor finding Power, that's where I'll use 13kJ / 37,83 sec. right? = 343.64 watts.

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