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Revision as of 14:39, 21 May 2010 (edit) Ian (Talk | contribs) (New page: We first obtain the acceleration <math>a</math>. Using the kinematic equation (see section 6) <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> <math>\begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...) ← Previous diff		Current revision (14:45, 21 May 2010) (edit) (undo) Ian (Talk | contribs)
Line 4:		Line 4:
	<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>		<math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math>
		+
		+	gives
	<math>\begin{align}		<math>\begin{align}
	& {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\		& {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\
-	& \\	+	& a=-\frac{2500}{600}=-4\textrm{.}167\ \text{m}{{\text{s}}^{-2}} \\
-	& a=-\frac{2500}{600}=-4.167\ \text{m}{{\text{s}}^{-2}} \\	+
	\end{align}</math>		\end{align}</math>
		+
		+	Then <math>F=ma\ </math> gives
		+
		+	<math>F=1000\times 4\textrm{.}167=4167\ \text{N}</math>

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Current revision

We first obtain the acceleration \displaystyle a.

Using the kinematic equation (see section 6)

\displaystyle {{v}^{\ 2}}={{u}^{\ 2}}+2as

gives

\displaystyle \begin{align} & {{0}^{2}}={{50}^{2}}+2\times a\times 300 \\ & a=-\frac{2500}{600}=-4\textrm{.}167\ \text{m}{{\text{s}}^{-2}} \\ \end{align}

Then \displaystyle F=ma\ gives

\displaystyle F=1000\times 4\textrm{.}167=4167\ \text{N}