Solution 10.2
From Mechanics
(Difference between revisions)
(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...) |
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<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> |
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}