Ian: New page: In this case there is resistance to the motion. Energy at start = Energy lost + Energy and finish. $\begin{align} & 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\t...$

Ian — Mon, 04 Oct 2010 12:12:44 GMT

New page: In this case there is resistance to the motion. Energy at start = Energy lost + Energy and finish. <math>\begin{align} & 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\t...

New page

In this case there is resistance to the motion.

Energy at start = Energy lost + Energy and finish.

<math>\begin{align}
& 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\times 40\times {{8}^{2}} \\
& \text{Energy Lost }=\text{ 3424 J} \\
\end{align}</math>

Solution 17.9b - Revision history

Ian: New page: In this case there is resistance to the motion. Energy at start = Energy lost + Energy and finish. \begin{align} & 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\t...

Ian: New page: In this case there is resistance to the motion. Energy at start = Energy lost + Energy and finish. $\begin{align} & 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\t...$