Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.
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cross-country skier is going up a slope at angle 5º to the horizontal. She is skating so only her skis provide the propulsion (like ice skating) (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?
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.
HELP PLEASE!!!
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You have written the second Newton's law for this problem just right!
But projections of forces are wrong.
First of all it is important to take into consideration what part of skier's weight falls on push leg. Let it be expressed by coefficient k=0.0 .... 1.0
F(static) = k*μs*m*g*cos(β)
F(kinetic) = (1-k)*m*g*cos(β)
Fg = m*g*sin(β), where β = 5º
k can be changed by the skier herself in order to gain the maximum acceleration.
It's obvious that the maximum acceleration is gained when F(net) is at its maximum.
This is achieved when k=1.0, that is the whole skier's weight falls on the pushing leg and the other one is up in the air.
Therefore F(net) = F(static)-Fg = μs*m*g*cos(β)-m*g*sin(β) = m*a
a = g*(μs*cos(β)-sin(β)) = 9.8*(0.12*cos(5º)-sin(5º)) = 0.32 m/s^2
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