Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Pretty please with a cherry on top?

  1. Mar 12, 2005 #1
    We're doing test corrections for my physics class. I did really well, however there were three questions that I had no idea how I was supposed to solve. here are the pesky critters that landed me an A-. :grumpy: :

    On this one, I guessed. I had and have absolutely no idea as to how in the world I'm supposed to do this. on the other two, I at least had some idea.
    so, can some one show me how to do this?

    thank you!!
    Last edited: Mar 12, 2005
  2. jcsd
  3. Mar 12, 2005 #2


    User Avatar
    Homework Helper

    For the first question, assume that the car goes at a constant velocity. So F = 850 N.
    How much work is done by the engine if the car goes for 17 km? (14.45 MJ)
    So the engine will do 14.45 MJ using 1 litre of gasoline.
    1 litre of gasoline can produce a 42 MJ work.
    So the efficiency is [itex]\frac{14.45}{42}[/itex]

    Do the same for number 3. You must multiple by 100 then devide by 27.

    For the second. You have already worked out the work done by gravity force (18228 J). So the work done by the cord must equal to the work done by gravity force to make that human have the same speed as he started jumped out of the bridge (i.e 0 m/s). So when [itex]W_{rope} = W_{gravity}[/itex] that human will stop going down and start going up.
    [tex]W_{cord} = \frac{1}{2}k(x_{2}^{2} - x_{1}^{2})[/tex]
    Were x1 and x2 is the difference between the length of the cord at the time measured and the length of the normal cord (12m)
    Note that when the man goes down 12m. The rope starts acting force. So x1 = 0m. And x2 = 31m - 12m = 19m.
    I get arround 100 N/m (not 10N/m as you said).
    Viet Dao,
  4. Mar 12, 2005 #3


    User Avatar
    Science Advisor

    VietDao29 has given good answers.

    I would point out that your error in problem 2 was assuming that the 31-12= 19 m stretch was the "x" in "F= kx". If the 60 kg person were just hanging from the bungee cord at a distance 19 m then that would be true- but in this situation, going down to 19m causes the person to bound back upward. Yes, you should use conservation of energy to find k. The person's kinetic energy at the bottom is 0 so the change in potential energy falling 31 m must be accounted for in the word done by the bungee cord in stretching 19 m (notice the difference in distances!).
  5. Mar 13, 2005 #4
    number three:

    how much work is required to accelerate a car with mass 2220 kg from 50.0 km/h to 80km/h in 10sec if its engine's efficiency is 27%?

    W=1/2 (mv2)2-1/2(mv1)2
    = 1/2 (2220 kg * [80.0 km/h]2)2 - 1/2 (2220 kg * [50 km/h]2)1
    =4329000 J= 4.329 MJ

    [(4.329 MJ)100%]/27% =16.03 MJ = 1.603 x 10^3 MJ

    So the work done on the rope is 18228J.

    I see. The rope's initial length is zero because that's where the initial measure ment is made.

    Wrope=1/2 k (x22 - x12)

    working equation:
    (2W)/(x22 - x21 = K

    2 (18228 J) / ([19m]2 - [0m]2) = 100.986 N/m = 1.0 x 10^2 N/m

    wow! thank you so much!!!

    So basically the elastic potential energy is going to equal the work done by gravity. THANX.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook