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!

Physics: Efficiency of toy car going upwards on a slope

  1. Jun 28, 2008 #1
    the toy car gains energy by pushin down the "head" of the "driver" and is powered by a spring that powers the wheels. at different heights, it has different spring constant. I mean the slope's angles are changed at different periods of the experiment. I am unable to work out the spring constant. isn't it supposed to be Work done by the spring = .5*spring constant*displacement^2?? for different angles, my spring constants are changing. any help in solving my mystery??
     
  2. jcsd
  3. Jun 28, 2008 #2
    No, that can't be right. The spring constant is something inherent in the spring that will always stay constant, unless of course it is pushed beyond its elasticity limit. I think you must be measuring the spring constant wrong.

    Remember that

    [tex]Work = \int \mathbf{F} \cdot d\mathbf{l}[/tex]

    Are you accounting for different forces at different angles acting on the spring?

    The equation you are using is correct because
    [tex]\frac{-d U}{dx}=F=-kx[/tex]
    which when you carry the dx to the other side and integrate between 0 and length s you will find that
    [tex]U(s)=U(0)+\frac{1}{2}ks^2[/tex]
     
  4. Jun 28, 2008 #3
    Thanx Mindscrape,
    however, da force i put in was the same. we used da one toy car and pressed it down till it cld not go down any further. how do i account for da different forces at different angles?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Physics: Efficiency of toy car going upwards on a slope
  1. Physics toy?~ (Replies: 5)

  2. Physics Toys (Replies: 7)

Loading...