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Homework Help: Energy and Momentum Questions

  1. Apr 10, 2004 #1
    1. A bicylist of mass 80 kg (including the bike) can coast down a 4 degree hill at a steady speed of 6.0 km/h. Pumping hard, the bicylist can descend the hill at 30 km/h. Assume the force of friction is proportional to the square of the speed; that is [tex]f_{fr} = bv^2[/tex], where [tex]b[/tex] is a constant.

    2. Two mountains are at elevations of 850m and 750m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one, with a total length of 3.2km and an average slope of 30 degrees,
    a) A skier starts from rest on the higher peak. At what speed will be arrive at the top of the lower peak if he just coasts without using poles?
    b)How large a coefficient of friction would be tolerable without preventing him from reaching the lower peak?

    1. no clue how to even start this one

    2a) (m)(9.8)(850) = (m)(9.8)(750) + (0.5)(m)([tex]vf^2[/tex])
    [tex]vf^2[/tex] = 1960
    [tex]vf[/tex] = 44.27 m/s

    b)I tried to solve this one...but it does not seem to be working.

    Thanks for your help. Much appreciated.
    *Ignore the _ that seems to be appearing after the variables. I don't know how to get rid of it.
    Last edited: Apr 10, 2004
  2. jcsd
  3. Apr 10, 2004 #2
    What do you need to find for Problem 1?
  4. Apr 10, 2004 #3
    Sorry..let me restate problem 1. Revision is in red.

    1. A bicylist of mass 80 kg (including the bike) can coast down a 4 degree hill at a steady speed of 6.0 km/h. Pumping hard, the bicylist can descend the hill at 30 km/h. Using the same power, at what speed can the bicyclist climb the hill? Assume the force of friction is proportional to the square of the speed; that is [tex]f_{fr} = bv^2[/tex], where [tex]b[/tex] is a constant.
  5. Apr 10, 2004 #4


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    1. start with this: he is moving down without using force at a steady speed of 6 km/h. component of force of gravity which pulls him forward = force of friction at that speed.

    (i got the result 28.8 km/h)
  6. Apr 10, 2004 #5
    I am finding the force of gravity as (80)(9.8)(cos4) = 782 N

    Since 6km/h = 1.667 m/s;

    Ffr = bv2
    782 N = bv2
    782 N= b (1.667 m/s)2
    b = 281

    Since 30km/h = 8.333 m/s;

    Ffr = bv2
    Ffr = (281)(8.333)2
    Ffr = 19514 N

    Now, i know that P = W/t = Fd/t = Fv

    P = Fv
    P = 19514(8.3333 m/s)
    P = 162614 Watts

    Now what can I do?
  7. Apr 10, 2004 #6


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    this is the component of the force of gravity that pulls him to the ground, you need the one that pulls him forward.

    btw, disregard the 28.8 solution i think i screwed up somewhere :/
  8. Apr 10, 2004 #7
    How do you find the force of gravity that pulls him forward?
    Could I please see what you did? It might help me as I am stuck.
  9. Apr 10, 2004 #8


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    i don't really know anything about physics so excuse my omitting of units..

    G = 9.81 * 80 = 784.8
    v1 = 1.666 m/s

    Gf = G * sin 4 = 54.74 (it is not large because the slope is not very steep)
    Gf = Ffr
    Gf = b * v1^2


    v2 = 8.333 m/s

    Ffr = b * v2^2
    Ffr = 1368.75

    he is moving at steady speed so forces are in balance, his own force (Fb) and gravity pushing him forward, and friction stopping him:

    Fb + Gf = Ftr
    Fb = 1314

    P = 1314 * 8.333 = 10950 (?)


    v3 = ?
    Gf = 54.74

    Fb = Ftr + Gf (he is going up now so gravity is stopping him)

    now, i got the original result by misreading "using the same power" as "using the same force". if it were "force", you would just insert Fg, Fb, and b and get v3 from this (8 m/s).

    if using the same power means that force he is using * his speed must be equal, then:

    10950/v3 = 19.71v3^2 + 53.74
    19.71v3^3 + 53.74v3 = 10950
    v3 = 8.11 m/s

    but i am not sure.. this "equal power" thing seems strange to me when gravity is also involved in the speed.. but i don't know physics, i'd like someone who does to solve this so i can be sure :)
    Last edited: Apr 10, 2004
  10. Apr 10, 2004 #9
    I understand what you did to get up to here:
    Ffr = 1368.75.

    However, from then on, I am confused. Thanks for your help so far.
  11. Apr 11, 2004 #10

    Doc Al

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    Staff: Mentor

    bicyclist problem

    Here's how I would analyze this problem. You have 3 unknowns: the force F that the bicyclist exerts when pumping hard, the constant "b" in the friction force formula, and the speed V going up the hill.

    Luckily you have 3 equations, obtained from applying the condition for equilibrium (total force is zero) to each of the three situations.
    (1) coasting downhill (V = 6 km/hour) allows you to find b:
    mg sinθ - b V2 = 0
    (2) pumping downhill (V = 30 km/hour): mg sinθ - b V2 + F = 0
    (3) pumping uphill: -mg sinθ - b V2 + F = 0

    Now solve for V (uphill).
  12. Apr 12, 2004 #11


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    looks like "using the same power" probably just means that he is exerting the same force, my original result is correct and i was just being stupid :(
  13. Apr 12, 2004 #12
    ugly quadratic

    how do u solve from the v^3 + v stage.. i probably should know this but i dont :P thanks...
  14. Apr 12, 2004 #13


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    Skiboka, that part is wrong and doesn't make sense, i apparently misunderstood the question.

    But if you want to know how to solve equations like that anyway:

    You have an equation in the form mx^3 + nx = o (m, n and o are some numbers).

    Divide by m, x^3 + nx/m = o/m. Let p=n/m, q=o/m. Now you have an equation in the form of x^3 + px + q, where p and q are some numbers.

    (a-b)^3 + 3ab(a-b) = a^3 - b^3 (you can easily check this)

    Let x=a-b, we get:
    a^3 - b^3=q

    a^3 - p^3/(27a^3) = q / * a^3
    a^6 - qa^3 - p^3/27 = 0

    t = a^3
    t^2 - qt - p^3/27 = 0

    This is a quadratic equation, find t, get 3rd root, that is a. b=p/(3a). x=a-b.
  15. Apr 12, 2004 #14
    i got to the final equation easily but how did u get v3 in one step???
  16. Apr 12, 2004 #15
    Ok...in response to the three equations..They make sense, but they assume that force is equal.. The answer I get is 8 m/s. The question asks if "the cyclist uses the same power, at what speed can he climb the hill"

    Does using the same power mean using the same force.

    I know that P = Work / time = Fd/t = Force*Velocity

    So, how do i find the answer? Thanks.....i really want to gain a good understanding of this problem. Could anyone on this board help me?
  17. Apr 12, 2004 #16
    You know the force, and you know the velocity, so... Look at your equation!

  18. Apr 12, 2004 #17
    Ok, i know the Force is 1312.67N
    and the velocity is 8.33 m/s

    the eqn for climbing the hill is mgsin4 + bv2 = Fperson

    i need to find the new v for going uphill.

    i know what mgsin4 and i know what b is....what if Fperson?
    I need to know Fperson to find the v going uphill.

    Last edited: Apr 12, 2004
  19. Apr 13, 2004 #18


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    By using the method I explained in my previous post. 19.71v3^3 + 53.74v3 = 10950 is of the form mx^3 + nx = o. I didn't get it in one step, I was just to lazy to copy everything from the paper. :)

    But as I said, that solution is wrong (for the question, it is correct for the equation itself), the right one is 28.8 km/h.

    st3dent: Your "Fperson" is what Doc Al named "F" in his equations. There are only 3 important forces here - gravity's component mgsin4, the force of friction bv^2, and the force that the bicyclist is exerting F. You can easily get b from the first equation, F from the second, and uphill speed from the third, assuming "using equal power" means "exerting the same force".
    Last edited: Apr 13, 2004
  20. Apr 13, 2004 #19

    Thanks for helping me.
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