- #1

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y=ax^2+bx+c, it has a y intercept of h and a repeated root l. a>0. The mass slides from x=0 to x=l with no friction.

[itex] d\vec{r} = \left(

\begin{array}{c}

1 \\

2ax+b

\end{array}

\right)\text{dx} [/itex]

A unit vector in this direction:

[itex] \hat{u} = \frac{1}{\sqrt{1+(2ax+b)^2}}\left(

\begin{array}{c}

1 \\

2ax+b

\end{array}

\right) [/itex]

To get the size of the component of the force parallel to the curve:

[itex] mg \cos (\theta) = mg\frac{(2ax+b)}{\sqrt{1+(2ax+b)^2}} [/itex]

[itex] mg \cos (\theta) \hat{u} = mg\frac{(2ax+b)}{\sqrt{1+(2ax+b)^2}} \frac{1}{\sqrt{1+(2ax+b)^2}}\left(

\begin{array}{c}

1 \\

2ax+b

\end{array}

\right) = mg \frac{(2ax+b)}{1+(2ax+b)^2}\left(

\begin{array}{c}

1 \\

2ax+b

\end{array}

\right) [/itex]

I don't think this is right