physics question if you understand please help?

Question

I have a physics question...Please if you understand explain this to me....I do not want anyone to do this for me I need explanation....b/c I wil need to understand this stuff when i take teh test so please explain adn talk me through this problem....

A skier, starting from rest, coasts down a mountain that makes an angle 25 degrees with the horizontal. The coefficient of kinetic friction between the snow and skis is .200. she coasts for 8.4 m before coming to the edge of a cliff.. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.80 m below the edge. How fast is she going just before she lands?

It says that this correlates the the work energy theorem and kinetic energy ... i read the sections and example problems and still have no idea where to even begin. Please help....thank you

Answer 1

This problem has a lot of little wrinkles in it. Plainly, you can do it correctly, but you have to do it carefully, one step at a time. Without doing the math, I'll try to walk you through it.

1. Let the skier's mass be m. The vertical force Fv = ma = gm, where g = -9.8 m/s2, applies at time zero, when the skier is at rest. Break that into components -- one headed down the slope (p for parallel), and one normal (perpendicular) to the slope. The formulas are Fp = Fv sin 25 and Fn = Fv cos 25. Draw a diagram using the vertical, horizontal, and slope to satisfy yourself that these formulas are correct. Note that Fp^2 + Fn^2 = Fv^2; in other words, Fv, the resultant force, is the hypoteneuse of the force triangle.

[By the way, you have the skier's mass m in all these equations. Just carry that along as m, a constant. In the end, it will wash out.]

2. The skier goes down the slope, propelled by the parallel force Fp, but retarded by the frictional force Ff. Calculate the frictional force Ff = -(mu) Fn, where mu = 0.2 is the coefficient of kinetic friction. The negative sign indicates the frictional force points up the slope, since both Fp and Fn are negative. Also, at this point, you can probably start using absolute values, since you know the skier is going downhill all the way.

The net force propelling the skier down the hill is Fs = Fp - Ff, where s stands for slope, and all three forces are absolute values.

3. Next, get the skier's speed at the edge of the cliff. The general formula is s = (1/2)at^2 + vo t + so, where vo = so = 0, so s = (1/2)at^2. You have s = 8.4 and acceleration a = Fs/m. Solve that equation for t, the time it takes to coast to the edge of the cliff.

4. To get the speed at the edge of the cliff, use v = at, with the acceleration and time values from step 3. Notice that the direction of this velocity is along the slope, 25 degrees from the horizontal.

5. Now that we know the skier's velocity at the edge of the cliff, we can use the work-energy theorem for the rest of the problem. Her kinetic energy at the top of the cliff is Kt = (1/2)mv^2. As she falls through the vertical distance 3.80 meters under the influence of gravity, the (negative) work done is W = Kb - Kt (where t is top and b is bottom) = mgh = 3.80 mg.

6. Therefore Kb =Kt + 3.80mg = (1/2)mv^2, where this v is the speed you're looking for -- the skier's speed when she lands.

This method will work. I should also note that there may be an easier way, using work-energy from the very beginning. The skier's kinetic energy at the top of the slope is zero, since she's not moving. The total vertical distance to the top of the cliff is 8.4 sin 25. You can work out the net vertical force acting on the skier on the slope by resolving the frictional force into horizontal and vertical components, then applying the work-energy theorem from the top of the slope.

There's a chance that might be easier, but if I were doing this problem myself, I'd do it the way I outlined in steps 1 through 6.

Answer 2

The first thing you need to do is figure out how fast the skier is going when she reaches the cliff. To do this you must calculate the force of friction and the force in the downhill direction. That will give you a net accelleration .
(F=Ma --> Fdownhill-Ffriction=Ma)
Find a formula that uses distance and acceleration to give you the velocity after 8.4 m. Remember initial velocity is 0 m/s. Once you have the velocity at the cliff you need to break it up into its vertical and horizontal components. Then assuming no air resistance you calculate acceleration of her falling body and use the same equation you used in the first part to get a final vertical velocity after 3.8 m. recombine the horizontal velocity(its unchanged) with the new vertical velocity to get final velocity. note If you have trouble with the trigonometry and calculating forces on angles then you probably need to ask another question to get help with vectors

Answer 3

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