Coefficient of Friction Needed?

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The discussion revolves around a physics problem regarding the horizontal force required for an 80-kg skier moving at constant velocity on snow. The key point is that the problem lacks information about the coefficient of friction, which is essential for calculating the applied force. It is suggested that the surface may be nearly frictionless due to the waxed skis, yet the answer key indicates a required force of 40N, implying a coefficient of friction of 0.05. Participants agree that the information provided in the question is insufficient for a definitive solution. The conversation highlights the challenges of interpreting physics problems with incomplete data.
haganjp
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Homework Statement


I was reviewing some questions from the NYS Regents Physics exam and came across this question, which I could not answer. Here's the question ...

An 80-kg skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles. What is the horizontal component of the force pushing him forward?

Homework Equations



In the horiztonal direction ...

Sum of Forces = mass x acceleration = 0 (constant velocity, therefore a=0)
Sum of Forces = Applied Force - Friction Force = 0

Applied Force = Friction Force
where, Friction Force = coefficient of Friction x Normal Force

Normal Force = m x g = 80kg * 9.81 m/s^2 = 784.8 N


The Attempt at a Solution



It seems I cannot determine a solution unless I know what the coefficient of friction is (which was not given in the problem statement). Am I missing something here?

It seems that it is implied that the surface is frictionless, in which case there is no need to apply a force to keep moving at a constant velocity, yet the answer key states that 40N of force are required. If this is indeed the answer, then a coefficient of friction of 0.05 is implied.

Thanks,

Jim
 
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Yes, seems information given in question is insufficient.
 
Hi there,

From the information you gave in this problem, there is really an information missing. In my understanding also, the idea of the wax on the skis reduces the friction coefficient to a minimum.

Then again, \mu = 0.05 is not a very big coefficient, which could definitely be right.
 
For the Regents exam... somehow I'm not too surprised :-/
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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