Solution 17.6c - Revision history

Ian at 12:20, 2 October 2010

2010-10-02T12:20:01Z

←Older revision		Revision as of 12:20, 2 October 2010
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-	[[Image:17.6a.gif]]	+	[[Image:17.6b.gif]]

	Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined.		Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined.

Ian at 12:17, 2 October 2010

2010-10-02T12:17:59Z

←Older revision		Revision as of 12:17, 2 October 2010
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		+	[[Image:17.6a.gif]]
		+
	Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined.		Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined.

-	The initial potential energy has been calcukated in part a) giving	+	The initial potential energy has been calcukated in part a) giving,


	<math>\begin{align}		<math>\begin{align}
-	& \text{Final }PE=5\times 9\textrm{.}8(8-8\cos 28{}^\circ ) \\	+	& \text{Final }PE=5\times 9\textrm{.}8(8-8\cos 28{}^\circ ) =45\textrm{.}9\text{ J} \\
-	& =45\textrm{.}9\text{ J} \\	+
	& \text{Energy Lost}=\text{52}\text{.5-45}\text{.9}=\text{6}\text{.60 J}		& \text{Energy Lost}=\text{52}\text{.5-45}\text{.9}=\text{6}\text{.60 J}
	\end{align}</math>		\end{align}</math>

Ian: New page: Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined. The initial potential energy has been calcukated in ...

2010-10-02T11:57:20Z

New page: Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined. The initial potential energy has been calcukated in ...

New page

Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined.

The initial potential energy has been calcukated in part a) giving

<math>\begin{align}
& \text{Final }PE=5\times 9\textrm{.}8(8-8\cos 28{}^\circ ) \\
& =45\textrm{.}9\text{ J} \\
& \text{Energy Lost}=\text{52}\text{.5-45}\text{.9}=\text{6}\text{.60 J}
\end{align}</math>