How Do You Calculate the Stopping Time and Displacement of a Skier on a Slope?

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To calculate the stopping time and displacement of a skier on a slope, one must consider the forces acting on the skier, including friction and gravity. The skier's initial velocity is 3.5 m/s, with a mass of 25 kg and a slope of 5 degrees, alongside a friction coefficient of 0.20. The net force can be determined using Newton's Second Law, leading to an acceleration that accounts for both gravitational and frictional forces. The stopping time was initially calculated as 1.3 seconds, and displacement was found to be 23 meters, but these results were questioned as potentially incorrect. A suggestion was made to approach the problem using kinetic energy concepts, indicating a need to reassess the calculations based on the forces involved.
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Homework Statement


A young skier (25kg) pushes off with ski poles to give herself an initial velocity of 3.5 m/s down a hill with 5o slope with a coefficient of friction of 0.20.
Find the time until skier comes to stop and her displacement.
Therefore
m = 25kg
vi = 3.5m/s
\theta = 5o
\mu = 0.20

Homework Equations


All kinematic-based equations
Newton's Laws (specifically \stackrel{\rightarrow}{F} = m(\Delta\stackrel{\rightarrow}{v}/\Deltat)
FF = \muFN


The Attempt at a Solution


My idea for this was to find the FNET and sub that into Newton's Second Law, solving for \Deltat (assuming that \stackrel{\rightarrow}{v}f is zero)
My solution for \Deltat was 1.3 seconds, correct to two significant digits (to me, that seems quite off)

For the second half of the question, finding the distance, I used \Delta\stackrel{\rightarrow}{d} = \stackrel{\rightarrow}{v}i + \frac{1}{2}\stackrel{\rightarrow}{a}(\Deltat)2 using \stackrel{\rightarrow}{a} as the component force of gravity acting parallel to the hill and I got 23m, correct to 2 significant digits.

I was then told I completed the question completely wrong. Now I'm lost. What was I supposed to do?
 
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Welcome to PF.

Maybe consider it is a Kinetic Energy to work from friction problem?

Work would be the (μ*mg*cos5° - mg*sin5°) times the distance and that would equal the KE.
 
I haven't learned kinetic energy yet...

BTW, this is a correspondence coarse for Grade 12 Physics in Ontario..
 
You know the frictional force, so find the net acceleration on the skier due to friction and gravity.
 
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