Tough problem (for me at least )

  • Thread starter benji
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  • #1
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Alright, so here's the problem:

An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25.0-degrees with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands?

So I didn't want to explain everything I tried, so I just scanned my paper ;). Here it is:

http://img129.exs.cx/img129/6093/physics_problem.gif

Can anyone see what I'm doing wrong? The correct answer is supposed to be 10.9 m/s...
 

Answers and Replies

  • #2
184
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Don't use forces, use energy

the velocity whe the skier leaves the cliff is...
[tex]\frac{1}{2}mv^2 = mgh[/tex] while h = 10.4m sin 25
v = sqrt(2g(10.4 sin 25)) = 9.29 m/s

the skiers velocity is 25 degrees below the horizontal, thus
[tex]v_x = v cos 25 \ v_y = v sin 25 [/tex]
thus solve for the time to fall the height 3.5 m
[tex] 0m = 3.5 m - v sin 25 t - 1/2 g t^2[/tex]
t= 0.53s

and when the skier makes impact the velocity is
[tex]v_y = v sin 25 + 9.8 (0.53 s)[/tex]
[tex]v_x = v cos 25[/tex]
so the total velocity at impact is..
[tex] v = sqrt( (v_x)^2+(v_y)^2) = 12.4128 m/s[/tex]

so i believe your answers wrong doing this is also very complicated but i thought would make u believe the answer more cause I use projectiles, gravity is a conservative force so therefore it doesn't matter what path the skier takes to drop the height u can find the velocity by 1/2mv^2 = mgh
where h is (10.4m sin 25 + 3.5m) the total height from the ground. you'll get the same result.

edit: i forgot the coff of friction, however use work engery principle in the same way i did this problem that might help
 
Last edited:
  • #3
382
2
Your sum of all x component forces :
[tex]F_{g}sin25-F_{fricition}=ma_x[/tex]
Your sum of all y component forces :
[tex]\Sigma F_{y}=ma_y[/tex]
N-mgcos25=0
 
Last edited:
  • #4
48
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Thanks guys, really appreciate your help.
 

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