1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?



    Sol:
    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

    pig

    User Avatar

    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

    pig

    User Avatar

    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

    pig

    User Avatar

    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
    b=19.71

    -------------------

    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

    User Avatar

    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

    pig

    User Avatar

    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

    pig

    User Avatar

    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:
    3ab=p
    a^3 - b^3=q

    Then:
    b=p/(3a)
    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!

    cookiemonster
     
  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.

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

    pig

    User Avatar

    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

    Thanks for helping me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Energy and Momentum Questions
Loading...