Velocity of skier slides off a frictionless hill

[Sucks@Physics]

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 don't know how to compute this without the angle

 

[Doc Al]

What's conserved?

 

[Retsam]

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]

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.

 

[Sucks@Physics]

The books says the answer is 20m/s?

 

[Doc Al]

Use conservation of energy, but be sure to apply it correctly.

 

[Sucks@Physics]

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]

E = 60*9.81*20 = 11772 J

That's the increase in KE as the skier goes down the hill. What's the initial KE? Final KE? Final speed?

 

[Sucks@Physics]

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?

 

[fliinghier]

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.

 

[Sucks@Physics]

sweet, thanks