A ski starts from rest and slides down a incline 85 m long.
- If the coefficient of friction is 0.090, what is the ski’s speed at the base of the incline?
- If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.
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This is Giancoli Answers with Mr. Dychko. The skier starts from rest and goes down this incline, a distance, d, and the incline has a coefficient of friction, µ. And there is friction force acting straight up the slope, there's a normal force perpendicular to the slope and gravity is downwards and I have the axis tilted so that x-axis is along the slope, positive downwards. So the initial potential energy equals the final kinetic energy that's down here plus the energy dissipated by friction. So this kinetic energy is gonna be less than the initial potential energy so we have to add this compensating thermal energy term in order to make this total equal to the starting total. And we can solve for the final kinetic energy by subtracting the energy dissipated by friction from both sides and we get final kinetic energy is initial potential minus the force of friction times distance. And that's one-half mv f squared; final kinetic energy equals mgh minus force of friction times d. And then we can solve for v f by dividing every term by m. So that's force friction times d over m and then multiplying everything by 2. So we have final speed then is square root of 2gh minus 2 times force of friction times d over mass. Now, we can't solve this equation because we don't know what the force of friction is yet so that's the next thing we turn our attention to. Force of friction is µ times normal force and the normal force is going to equal the y-component of gravity because there's no acceleration perpendicular to the slope; this acceleration is down the slope. So that must mean that the forces that are perpendicular to the slope balance each other. So, the normal force, on the one hand is equal to the y-component of gravity, on the other hand and this is a force of gravity mg times cos Θ because it's the adjacent leg of the gravity-vector triangle. And so here we have normal force, y-component of gravity, mgcos Θ and we substitute mgcos Θ, in place of F N here, to get the friction force is µmgcos Θ. The height that the person falls is because we need to substitute for h here and because we know what d is so we need to rewrite h in terms of d. h is gonna be d times sin Θ because this vertical height is the opposite leg of this triangle here and d is the hypotenuse. So we use hypotenuse times sin Θ to get the opposite h. So, we'll substitute in dsin Θ for h here and we'll substitute in µmgcos Θ for force of friction here and we rewrite our velocity formula now. It's gonna be square root 2 gdsin Θ minus 2µmgcos Θ times d over m. And we have 2gd is the common factor so we will factor that out to make our writing a little bit simpler; we have final speed is 2gd times sin Θ minus µcos Θ all square rooted. And so that's the square root of 2 times 9.8 meters per second squared times 85 meters—distance along the slope— times sin 28—angle of incline to the slope— minus 0.09—coefficient of friction— times cos 28 that gives 25 meters per second will be the final speed after accounting for the loss of energy due to friction dissipated as thermal energy. And then once they reach the bottom of the slope, the question is, how far will they go? And we have that the initial kinetic energy, which is kinetic energy here, is gonna be dissipated entirely into heat by this by the friction force and it will do it will turn an amount of energy equal to the friction force times the distance, x over which the force acts into thermal energy. So we have one-half mv initial squared equals force of friction times x. And we'll solve for x by dividing both sides by force of friction. And we know the force of friction is µF N and in this case, our free-body diagram is a little simpler because the gravity force upwards or sorry, normal force upwards equals the gravity force downwards and there's no angles to consider here. So, normal force is just mg now. So, we substitute in µmg for the friction force here and we get then that the distance will be v initial squared over 2µg; the m's cancel there, on top and bottom and so x is gonna be 25.4902, which we figured out from part (a). And let's square that speed divided by 2 times 0.09 times 9.8 and we get 370 meters is the total distance traveled.
how did you get 4902 toward the final the solution
i hope to hear from you
thanks
ebe
Hi anochc, thanks for the question. I think you're referring to when I said "25.4902 which we figured out from part 'a'" at the point 5:10 in the video. What I'm doing is substituting the answer from part "a" (twenty five point four nine eight zero two", for the initial velocity at the bottom of the slope, into the formula for distance in part "b".
Hope that helps,
Mr. Dychko
Whenever you do a triangle within the free body diagram, how do you know in which of the three corners to place the angle theta? Where you place the angle influences if it's going to be sin or cos, so how do you choose where? For example in this question and questions 5 and 10. Thank you.
Hi nlt1307,
Thank you for your question. It actually doesn't matter where you choose to put the angle. Your choice, as you say, determines which trigonometric function you'll use to find components, but there's no "standard". Looking at questions 5 and 10, it's convenient to label the angle between the perpendicular to the ramp and the vertical, since this is the same as the incline angle. Even though the triangles are drawn differently in problems 5 and 10, you'll notice theta is still between the perpendicular and the vertical.
Hope this helps,
Mr. Dychko
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