Differences
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| Both sides previous revision Previous revision Next revision | Previous revisionLast 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 04:39] – pwirving | ||
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| === Facts ==== | === Facts ==== | ||
| + | |||
| + | Block is metal. | ||
| Mass of metal block = 3 kg | Mass of metal block = 3 kg | ||
| Line 11: | Line 13: | ||
| 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 $ | ||
| - | Since the net force was constant, $v_{x,avg} = (v_{xi} + v_{xf})/2$, so | + | Since the net force was constant |
| $ \Delta x/\Delta t = ((6 + 0)/2) m/s = 3m/s $ | $ \Delta x/\Delta t = ((6 + 0)/2) m/s = 3m/s $ | ||
| + | |||
| + | Sub in for $\Delta t$ and solve for $\Delta x$ | ||
| $ \Delta x = (3 m/s)(1.53 s) = 4.5m $ | $ \Delta x = (3 m/s)(1.53 s) = 4.5m $ | ||