Coefficient of kinetic friction...(physics)

[RoyalXBlood]

Kieran takes off down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 400 N. The combined mass of skis and Kieran is 50 kg (the fuel mass is negligible). Kieran's speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow?

 

So far my work shows...

 

I know that force due to mass and gravity is 490. I'll call it [math]F_{g}[/math]. I found it's x and y components to be [math]F_{xg}[/math]= 85.1 and [math]F_{yg}[/math]= 482.6. So [math]F_{N}[/math] is 482.6. That gets me one step closer because I know I need to solve [math]F_{k}[/math]=[math]M_{k}[/math][math]F_{N}[/math]. I don't know how to find the force of kinetic friction.

 

I hope I wasn't confusing.

[big314mp]

What you will need to do is find the total force accelerating Kieran (who comes up with these names anyways?) down the hill. To do this, you will add the force of the skies, plus the Fxg force you calculated. That gives you the force applied to kieran.

 

Now, you know how long the slope is (well...you'll have to calculate it), so you know the total distance that kieran traveled. You know his final speed (40 m/s) and his initial speed (0 m/s). You can now calculate his acceleration down the slope.

 

You know kieran's mass, so using your calculated acceleration, you can figure out the force on kieran that is actually doing the acceleration.

 

Hopefully you can get it from there.

[RoyalXBlood]

I assume the equation I can use is [math]V^{2}[/math] = [math]V_{o}^{2}[/math] + 2a[math]\Delta[/math]x. From that I get a=15.75692405 (I like the keep all the figures). So the force accerating him is 787.8462075 Newtons. I still don't see how I can use this to find Force of kentic friction.

[big314mp]

Your calculation of acceleration is wrong. Correct equation, wrong (delta)X.

 

If you imagine the hill, it is a right triangle. The "up and down" leg of the triangle is 50m. Your (delta)X for the calculation needs to be the hypotenuse of this triangle.

 

Once you know how much force is actually accelerating the skier, then you can figure out how much force is lost to friction. Remember that the force applied to the skier (what is calculated in the first paragraph of my last post) is equal to the force of friction plus the force of acceleration. So by calculating the force of acceleration and the applied force, you can calculate the force of friction.

[RoyalXBlood]

Ok say leg b goes along the yaxis and leg a goes along the x-axis. C is the hypotenuse. B=50m. The only thing that I am not certain of is the angel, the given angel is that the angel formed between b and c or a and c?

 

Ok using 50/cos80 I get 287.29385242m, that must be it, then the force would be 138.9185421. The difference is 346.169065N. That would be [math]F_{k}[/math]. So the coefficient would be .7173658791 or just .72.

 

It's hard for me to understand because you basicly subtract the final force acting on the guy from the initial and that would get you the force of kenetic friction. I just don't understand why that is how it works.

Edited September 28, 2008 by RoyalXBlood
multiple post merged

[big314mp]

The angle of elevation will always be between the x-axis and the hypotenuse.

 

It's not "final" force and "initial" force. If the skier were accelerating on a frictionless surface, you could calculate the force acting on him based on his mass and acceleration. To achieve a given acceleration, will require a given force.

 

Now transfer that over to a surface with friction. All of the forces are the same, except there is one new one. This is the force of friction. The frictional force will act in the opposite direction that the skier is going; in other words, it will act to slow him down. This also means that it is acting in an opposite direction from the force accelerating the skier. To achieve the same acceleration, we now need to add additional acceleration force to counteract the frictional force holding the skier back. The amount of force we need to add (remember we are adding to the acceleration force calculated on a frictionless surface) is the same amount as the frictional force - in other words, enough force to cancel out the frictional force. So the sum of the forces accelerating the skier on a friction surface will be (force of acceleration) plus (force needed to counter friction). So if you know the total force applied, you need only subtract out the force of acceleration (on a frictionless surface) to calculate the frictional force.

[big314mp]

The top part shows a no friction example, the bottom shows a with friction example.

 

The black arrow is calculated by using F=ma. The total force applied is the black arrow plus the blue arrow. The red arrow is the force of friction. The blue arrow is the same magnitude as the red arrow. So to calculate the length of the blue arrow (and therefore the red arrow), you need to subtract the black arrow from the total force applied.

[RoyalXBlood]

Thank's for helping me out so much, I appreciate the dedication.

[drufae]

you could make the problem simpler by shifting your axes.i.e. take the components of all the forces along either the slope or perpendicular to it.

At the end you should get an f.b.d of the block some what like this:

 

giving the equations

[math]f_k = \mu mgcos \theta;

F + mgsin\theta - f_k = ma [/math]

 

shifting axes usually works really well in most kinematics problems.