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Velocity of skier slides off a frictionless hill

a Skier of mass 60 kg pushes off the top of a frictionless hill with an initial speed of 5.0 m/s. How fast will she be moving after dropping 20m in elevation?

i dont know how to compute this without the angle
 

Doc Al

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What's conserved?
 
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Use potential energy formula :

E = m*g*(change in height)

E = 60*9.81*20 = 11772 J

So skier loses 11772 J of potentiaql energy on descent, this is all converted into kinetic energy. So using KE formula :

KE = 0.5*m*(v^2)

V = sqrt(KE/0.5*m) = 19.81 m/s

So add the initial velocity to this to get :

5 + 19.81 = 24.81 m/s
 

Doc Al

Mentor
44,681
1,000
Use potential energy formula :

E = m*g*(change in height)

E = 60*9.81*20 = 11772 J

So skier loses 11772 J of potentiaql energy on descent, this is all converted into kinetic energy. So using KE formula :

KE = 0.5*m*(v^2)

V = sqrt(KE/0.5*m) = 19.81 m/s

So add the initial velocity to this to get :

5 + 19.81 = 24.81 m/s
(1) Please reread the forum rules about posting complete solutions.
(2) This solution is not correct.
 
The books says the answer is 20m/s?
 

Doc Al

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44,681
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Use conservation of energy, but be sure to apply it correctly.
 
E = 60*9.81*20 = 11772 J
KE = 0.5*m*(v^2)

I'm not exactly sure how to manipulate the KE formula to where it will come up with a reasonable answer
 

Doc Al

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initial = 750J
final = 12522J

so v^2 =12522J/(.5*m) = sqrt 417.4 = 20.4

And do you not add the initial push off speed since you used it to get the initial KE?
 
no you have already accounted for it since you used KE+PE at the beginning and got total KE at the bottom. basically because you had the KEi in the equation it's accounted for.
 

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