Time for Skier to Stop in 8.7° Slope: No Friction

In summary: Her kinetic energy is unchanged, so her potential energy is also unchanged. So her y-component of velocity would be 14.9 m/s, and her x-component would be 2 m/s.In summary, a skier with mass 65.9 kg is going 14.9 m/s on a flat surface. She hits a slope with angle 8.7 degrees from the horizontal. It takes 10.0 sec for her to stop.
  • #1
Ryo124
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0
A skier with mass 65.9 kg is going 14.9 m/s on a flat surface. She hits a slope with angle 8.7 degrees from the horizontal. How long does it take for her to stop? Neglect friction.

I've tried breaking up her velocity into components. Also, found the force on her once she is on the slope with mgsin[tex]\theta[/tex]. I don't know if I needed to do this or what to do once I found these values.

I'm just not sure how to start this and what to do from there?
 
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  • #2
She starts up the slope with her full speed, so no components needed. You found the force on her, so what's her acceleration? The rest is kinematics.
 
  • #3
Personally I'd just use conservation of energy.

Find her kinetic energy ([tex]KE = 0.5 * mv^{2}[/tex]), find at what height h she would have the same amount of gravitational potential energy ([tex]GPE = mgh[/tex]).

Once you know how high she has to go, and at what angle she has to do it at, it's trigonometry.
 
  • #4
I assume that this:
Ryo124 said:
How long does it take for her to stop?
means that they want the time it takes for her to come to rest.
 
  • #5
I keep getting 5.02 sec. That's not right.

The answer is 10.0 sec. Can anyone walk me through this to show why this answer is correct?
 
  • #6
Show us exactly what you did.

What was the acceleration? What kinematic equation did you use to find the time?
 
  • #7
Never mind, I solved it.

I solved for her force parallel to the plane, F||, and set that equal to ma. Then, I solved for a.

Then, I used the equation: V=Vo + at to solve for t.

However, when I solve for t in the above equation, I don't get why her initial velocity, Vo, is 14.9 m/s and not her y-component of velocity, which would be about 2 m/s.
 
  • #8
When she hits the slope we assume no energy is lost. The slope just changes her direction, not her speed--she's still going full speed when she starts up the slope.
 

FAQ: Time for Skier to Stop in 8.7° Slope: No Friction

1. What is the formula for calculating the time for a skier to stop on an 8.7° slope with no friction?

The formula for calculating the time for a skier to stop on an 8.7° slope with no friction is t = √(2d/gsinθ), where t is the time in seconds, d is the distance in meters, g is the acceleration due to gravity (9.8 m/s^2), and θ is the slope angle.

2. How does the slope angle affect the time for a skier to stop with no friction?

The slope angle affects the time for a skier to stop with no friction because it determines the rate of acceleration due to gravity. The steeper the slope, the greater the acceleration and the shorter the stopping time. A flatter slope will have a lower acceleration and therefore a longer stopping time.

3. What assumptions are made when calculating the time for a skier to stop on an 8.7° slope with no friction?

When calculating the time for a skier to stop on an 8.7° slope with no friction, it is assumed that the skier is traveling in a straight line, there is no air resistance, and there is no friction between the skier's skis and the snow.

4. How accurate is the calculated time for a skier to stop on an 8.7° slope with no friction?

The calculated time for a skier to stop on an 8.7° slope with no friction is an idealized estimate and may not be entirely accurate in real-world situations. Factors such as the skier's speed, weight, and technique can also affect the stopping time.

5. Can this formula be applied to other slope angles and surfaces?

Yes, this formula can be applied to other slope angles and surfaces as long as there is no friction present. However, for more accurate calculations, other factors such as air resistance and friction may need to be taken into account.

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