Velocity of skier slides off a frictionless hill

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of a skier descending a frictionless hill. The skier starts with an initial speed and drops a certain height, prompting questions about the conservation of energy and the calculation of final velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the conservation of energy, questioning how potential energy converts to kinetic energy during the skier's descent. There are attempts to calculate the final speed using energy formulas, with some participants expressing uncertainty about the correct application of these concepts.

Discussion Status

The discussion is active, with various approaches being proposed to solve the problem. Some participants have provided calculations, while others have raised questions about the assumptions made and the correctness of the methods used. There is no explicit consensus on the final answer, and participants are engaging in clarifying the application of energy conservation principles.

Contextual Notes

There are indications of confusion regarding the initial kinetic energy and how it factors into the final calculations. Some participants have noted discrepancies between their results and those found in reference materials.

Sucks@Physics
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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
 
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What's conserved?
 
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
 
Retsam said:
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?
 
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
 
Sucks@Physics said:
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?
 
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?
 
  • #10
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.
 
  • #11
sweet, thanks
 

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