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Homework Help: Energy vs. Newton's Laws

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 degree frictionless slope. A strong headwind exerts a horizontal force of 200 N on him as he skies.

    Find his speed at the bottom of the slope using Newton's Law
    Find his speed at the bottom of the slope using Energy

    2. Relevant equations
    Conservation of Energy
    F = ma
    vf^2 = vi^2 + 2ad
    ME_i = ME_f

    3. The attempt at a solution

    I set up a drawing, found the length of the slope, did the force analysis, and got the following (Newton's Laws part):

    Length of the slope = 146.19m
    Force down the slope = m*g*sin(20) = 251.3848N
    Force of wind up the slope = 200/cos(20) = 212.835

    So Fnet = 251.3848-212.835= 38.549
    a = Fnet/m = 38.549/75 = .5139


    vf*vf = vi*vi + 2(a)(d)
    vf*vf = 0 + 2(.5139)(146.19)
    vf = sqrt(150.28) = 12.25 m/s

    Not sure where I went wrong.

    As for the energy part, I'm not sure how to approach it.
  2. jcsd
  3. Apr 2, 2010 #2
    I would like help on this one too....(just out of interest). (A total amateur, I am)

    I tried it out in the energy way using eqns:
    mgh = 1/2 mv^2
    And also got an answer.
    Though by the rules of the forum I cannot tell u the answer.
  4. Apr 2, 2010 #3


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    For one thing, how can the force of wind up the slope be greater than the total horizontal force? And for The legend, mgh=(1/2)*mv^2 is only going to apply if the only force is gravitational.
    Last edited: Apr 2, 2010
  5. Apr 3, 2010 #4
    I know I deconstructed the horizontal force wrong, but geometrically I can't figure it out. Every time I try to draw the forces I get the same thing.
  6. Apr 3, 2010 #5


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    I get a right triangle with a 200N force along the hypotenuse and the adjacent side to a 20 degree angle being the force up the slope.
  7. Apr 3, 2010 #6
    Not sure if this helps, but the General Form of the work energy principle is Wnon-conservative = [tex]\Delta[/tex]KE + [tex]\Delta[/tex]PE. I think the work done by non-conservative force is the work done on the skier by the wind.
  8. Apr 3, 2010 #7
    I was mistaken about what headwind meant when applied to the problem.

    Same basic concept as before:

    Fnet = ma
    Fnet = m(g)(sin20) - 200
    Fnet = 51.3848N = (75)(a)

    a = .6851

    vf^2 = vi^2 + 2(a)(d)
    vf^2 = 2(.6851)(146.19)
    vf = 14.15 m/s

    Where'd I go wrong?
  9. Apr 3, 2010 #8
    what is the energy that the wind exerts on Sam, throughout his journey down? Give you a hint: Energy = Work = Force * Distance. This energy is part of the energy-work equation that you already know.
  10. Apr 4, 2010 #9
    All of the above work was for the Newton's Laws part of the equation. I'm asking what I did wrong on THAT part.

    As for the energy:

    Wnet = Change in KE
    Wnet = Wgravity - Wwind
    Wnet = 75(9.8)(sin(20))*146.19 - 200*146.19
    Wnet = 7511.944

    Wnet = KEf - KEi
    Wnet = KEf
    7511.944 = (1/2)(75)(v^2)
    v^2 = 200.318
    v = 14.1533

    Which is wrong.

    The headwind is horizontal, not going up the hypotenuse. I'm not sure how to deconstruct that force.
  11. Apr 4, 2010 #10


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    The force from the headwind is applied in a horizontal direction. You only want the component of that force PARALLEL to the slope. If you draw the force triangle for that 200N force and split it into components normal and parallel to the slope, the 200N force is the hypotenuse.
  12. Apr 4, 2010 #11
    So then the force acting up the hypotenuse is 200N times the sine of the angle?

    Sorry for all the questions. I can't get the picture right geometrically. On paper or in my head.
  13. Apr 4, 2010 #12


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    Draw the slope going down at 20 degrees. Draw the horizontal 200N force vector with origin on the slope. From the arrow end of the vector draw a PERPENDICULAR to the slope. NOT a vertical, which it looks like you been doing. If you've got it right the 200N force is the hypotenuse of a right triangle. The short vector leg of the triangle at an angle of 20 degrees to the vertical is the normal component to the slope and the longer at 20 degrees to the horizontal is the one you want. The component parallel to the slope.
    Last edited: Apr 4, 2010
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