Problem With Maximum Acceleration up Hill

  • #1
Problem With Maximum Acceleration up Hill!

Homework Statement

cross-country skier is going up a slope at angle 5º to the horizontal. She is skating so only her skis provide the propulsion (i.e. she does not push with her ski poles). The static and kinetic friction coefficients for this situation are μs = 0.12, μk = 0.07 respectively.

Find the magnitude of her maximum possible uphill acceleration?

Homework Equations


The Attempt at a Solution

So Far:

Fnet(x)= F(static) - F(kinetic) - Fg = ma

Fnet(x) = (0.12 x m x 9.8cos85) - (0.07 x m x 9.8cos85) - (9.8cos85 x m) = ma

Divide by m you get:

Fnet(x) = (0.12 x 9.8cos85) - (0.07 x 9.8cos85) - 9.8cos85 = a

Therefore a= - 0.81m/s^2

But this gives me a negative accelleration, Where did I go wrong? as it asks for the Max acceleration up the hill and that acceleration would point down hill.

  • #2

How can she possibly accelerate uphill?
  • #3

She uses her skis to push against the ground which propells her forward through static friction.
  • #4

I thought the question said "i.e. she does not push with her ski poles".
  • #5

she DOES NOT use her poles, but she DOES use her skis - the same way people ice skate
  • #6

Please do not double post.

You should first correct your geometry/trig/normal force calculation errors. Friction is a function of the normal force. The normal force is not mgcos85; what should it be?
Then, once you make your correction for the normal force equation, you must make certain assumptions about the distribution of the skiers weight ...that all of it is one one ski as she pushes back against the snow, causing static friction to propel her forward, while the gravity component downhill (which you have calculated correctly) is at the same time working against her. But what about the sliding friction on the other ski..what's the normal force on the other ski with all her weight initially shifted to the 'propelling' ski??
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