Solution 17.6c
From Mechanics
(Difference between revisions)
(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 ...) |
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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} \\ |
| - | + | ||
& \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> | ||
Current revision
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,
\displaystyle \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}
