Velocity of skier slides off a frictionless hill

In summary, the skier of mass 60 kg will be moving at a speed of 20 m/s after dropping 20m in elevation. This is because the skier loses 11772 J of potential energy on descent, which is converted into kinetic energy. The initial velocity of 5.0 m/s does not need to be added since it is already accounted for in the equation.
  • #1
Sucks@Physics
76
0
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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  • #2
What's conserved?
 
  • #3
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
 
  • #4
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.
 
  • #5
The books says the answer is 20m/s?
 
  • #6
Use conservation of energy, but be sure to apply it correctly.
 
  • #7
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
 
  • #8
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?
 
  • #9
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
 

1. How is the velocity of a skier sliding off a frictionless hill calculated?

The velocity of a skier sliding off a frictionless hill can be calculated using the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill. This formula assumes that there is no air resistance or friction acting on the skier.

2. What factors affect the velocity of a skier sliding off a frictionless hill?

The velocity of a skier sliding off a frictionless hill is affected by the height of the hill, the mass of the skier, and the acceleration due to gravity. Additionally, external factors such as wind resistance and the angle of the hill can also influence the skier's velocity.

3. How does the velocity of a skier sliding off a frictionless hill compare to a skier sliding down a hill with friction?

The velocity of a skier sliding off a frictionless hill will be greater than that of a skier sliding down a hill with friction. This is because friction acts as a resistive force, slowing down the skier's motion, while a frictionless surface allows the skier to maintain a constant velocity.

4. Can the velocity of a skier sliding off a frictionless hill ever reach zero?

No, the velocity of a skier sliding off a frictionless hill will never reach zero. This is because there is no resistive force acting on the skier, so they will continue to slide at a constant velocity until they reach the bottom of the hill or encounter another external force.

5. How does the angle of the hill affect the velocity of a skier sliding off a frictionless hill?

The angle of the hill has a direct impact on the velocity of a skier sliding off a frictionless hill. A steeper angle will result in a higher velocity, while a shallower angle will result in a lower velocity. This is due to the fact that the steeper the angle, the greater the force of gravity acting on the skier, increasing their acceleration and velocity.

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