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Skier Going Uphill - Intro Physics Problem

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data
    You've skied down a slope and are going 20m/s when you hit the base of a slope that is 30 degree incline.?

    Assuming friction and drag are negligible, how much altitude (y) will you gain as you’re slowing down?

    How far up the 30 degree slope (∆x along the incline) will you coast before you stop?

    I know the following variables:
    vi = 20 m/s
    vf = 0 m/s
    angle is 30°

    I know the skier's energy is kinetic changing to potiential gravitational energy.
    2. Relevant equations
    Ki → Ug

    KE = 1/2*m*v^2
    PE = m*g*h

    For a)
    It seems like I need to use 20tan30, but I'm not sure.

    For b)
    need ma = mgsin(theta) to solve for a = gsin(theta)

    and vf^2 = vi^2+2*a*d → d = -vf^2 + vi^2 + 2a


    3. The attempt at a solution
    For a) 20*tan(30°) = 17.32

    For b) (-9.8)sin(30°) = -4.9

    d = 0 + (20)^2 + 2(-4.9)
    d = 40 + (-9.8)
    d= 30.2
     
  2. jcsd
  3. Mar 22, 2012 #2
    For part A just consider the conservative forces. At the base of the hill the guy has a kinetic energy and no potential energy (relative to the base of the hill that is). When he goes up the 30 degree incline, eventually his velocity goes to zero (he stops) and his kinetic energy goes to zero. But, he will have gravitational potential energy.

    So what you can do is say that:
    [tex]\frac{1}{2}mv^{2} = mgh[/tex]
    That is, his kinetic energy before going up the 30 degree incline, is equal to his gravitational potential energy when he stops on the incline. You should have all the numbers to do this and solve for the h, which is his height (altitude).

    From there, for part B, just consider where he is at on the 30 degree right triangle. You know his height h (which becomes the y coordinate, and you know the degree of the incline, so using basic trigonometry you can find the hypoteneuse (the distance).
     
  4. Mar 22, 2012 #3
    Ok, that makes sense. Makes part B easy to find h in part A.

    So I want to rearrange [tex]\frac{1}{2}mv^{2} = mgh[/tex] to
    [tex] h = \frac{1}{2} \frac {mg}{mv^{2}} [/tex]

    So mass will cancel out, leaving me with the known variables. If that is the correct rearranged form, I know enough trig to solve for part b. Thanks.

    EDIT: I realized my rearranging is wrong.

    [tex] h = \frac{1}{2} g-v^{2} [/tex]


    [tex] h = \frac{1}{2} (9.8)-(20)^{2} [/tex] = 195.1 m

    Part B: 195.1sin(30°)= 97.55 m
     
    Last edited: Mar 22, 2012
  5. Mar 22, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Yikes! :surprised: How did you manage to work a new term into a proportional relationship?
    Also note that the units of g are not the same as the units of v2, so something MUST be wrong...

    Try again to rearrange the expression. What happens if you divide both sides by mg?
     
  6. Mar 22, 2012 #5
    I was rearanging it again last night since the new arrangement didn't seem right either.

    [tex] h = \frac{1}{2} \frac {mv^{2}}{mg} [/tex]

    I noticed that I had mv^2 on the bottom, but it should be on the top, while dividing by mg.
     
  7. Mar 22, 2012 #6
    Thats correct, so what's next?
     
  8. Mar 22, 2012 #7
    [tex] h = \frac{1}{2} \frac {400}}{9.8} = 20.4 m[/tex]

    Knowing that mass cancels out, I get 20.4 m = y

    If I use 20.4sin30 = 10.2. But I think that's too short.
     
  9. Mar 22, 2012 #8

    BruceW

    User Avatar
    Homework Helper

    You've got the right answer for y. But your answer for the distance travelled up the slope is not correct. Draw a diagram, so you can see how the lengths are related.
     
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