Solution 17.6c

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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.

Current revision

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.

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}