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Physics: Efficiency of toy car going upwards on a slope

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??
 
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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]
 
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?
 

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