Differences
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Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
183_notes:examples:sliding_to_a_stop [2014/09/16 07:51] – pwirving | 183_notes:examples:sliding_to_a_stop [2014/09/22 04:35] – pwirving | ||
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=== Facts ==== | === Facts ==== | ||
+ | |||
+ | Block is metal. | ||
Mass of metal block = 3 kg | Mass of metal block = 3 kg | ||
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The coefficient of friction between floor and block = 0.4 | The coefficient of friction between floor and block = 0.4 | ||
- | Initial velocity of block $\langle 6, 0, 0\rangle m/s$ | + | Initial velocity of block = $\langle 6, 0, 0\rangle m/s$ |
+ | |||
+ | Final velocity of block = $\langle 0, 0, 0\rangle m/s$ | ||
=== Lacking === | === Lacking === | ||
+ | |||
+ | Time it takes for the block to come to a stop. | ||
+ | |||
+ | The distance the block moves during this time. | ||
=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
+ | |||
+ | Assume surface is made of the same material and so coefficient of friction is constant. | ||
=== Representations === | === Representations === | ||
+ | |||
+ | {{183_notes: | ||
$\Delta \vec{p} = \vec{F}_{net} \Delta t$ | $\Delta \vec{p} = \vec{F}_{net} \Delta t$ | ||
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$ y: \Delta p_y = (F_N - mg)\Delta t = 0 $ | $ y: \Delta p_y = (F_N - mg)\Delta t = 0 $ | ||
- | Combining these two equations and writing $ p_x = mv_x $, we have | + | Write equation of y direction in terms of $F_N$ to sub into x direction equation. |
+ | |||
+ | $ (F_N - mg) \Delta t = 0 $ | ||
+ | |||
+ | Multiply out | ||
+ | |||
+ | $ F_N \Delta t - mg \Delta t = 0 $ | ||
+ | |||
+ | Make equal to each other | ||
+ | |||
+ | $ F_N \Delta t = mg \Delta t $ | ||
+ | |||
+ | Cancel $\Delta t$ | ||
+ | |||
+ | $ F_N = mg $ | ||
+ | |||
+ | Combining these two equations | ||
$ \Delta(mv_x) = -mg\Delta t $ | $ \Delta(mv_x) = -mg\Delta t $ | ||
- | $ \Delta(v_x) = - g\Delta t $ | + | Cancel the masses |
+ | |||
+ | $ \Delta(v_x) = - g\Delta t $ | ||
+ | |||
+ | Rearrange to solve for $\Delta t$ and sub in 0 - $v_{xi}$ for $ \Delta(v_x)$ | ||
$ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $ | $ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $ | ||
+ | |||
+ | Fill in values for variables and solve for $\Delta t$ | ||
$ \Delta(t) = \dfrac{6 m/s}{0.4 (9.8 N/kg)} = 1.53s $ | $ \Delta(t) = \dfrac{6 m/s}{0.4 (9.8 N/kg)} = 1.53s $ |