What Is the Maximum Acceleration a Cross-Country Skier Can Achieve Uphill?

AI Thread Summary
The discussion focuses on calculating the maximum acceleration a cross-country skier can achieve while skiing uphill at a 5º angle. The skier relies solely on her skis for propulsion, with static and kinetic friction coefficients provided. The user attempts to solve the problem using the equation Fnet=ma, arriving at an acceleration of 1.4 m/s². There is uncertainty about whether this calculation correctly accounts for the weight component due to gravity. The conversation highlights the importance of considering all forces acting on the skier to determine the accurate maximum acceleration.
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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.

what is the maximum accelleration she can obtain?

Homework Equations



Fnet=ma

The Attempt at a Solution



so far i have got to (0.17-0.02)*9.8 *cos85=a

which equates to 1.4m/s^2 is this correct?

I got it by dividing the 9.8*cos85 and the mass (which canceled out).

Juswanted to know if its right as i don't have the answer and been working on it for a while.
 
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What about her weight component (the gravity force)?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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