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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:sliding_to_a_stop [2014/09/22 04:33] – pwirving | 183_notes:examples:sliding_to_a_stop [2018/02/03 23:24] (current) – [Example: Sliding to a Stop] hallstein | ||
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| ===== Example: Sliding to a Stop ===== | ===== Example: Sliding to a Stop ===== | ||
| - | You take a 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the block with an initial velocity of $\langle 6, 0, 0\rangle m/s$. How long will it take for the block to come to a stop? How far does the block move? | + | You take a 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the block with an initial velocity of $\langle 6, 0, 0\rangle m/s$. How long will it take for the block to come to a stop? How far does the block move? |
| === Facts ==== | === Facts ==== | ||
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| === Solution === | === Solution === | ||
| - | $ x: \Delta p_x = -F_N\Delta t $ | + | $ x: \Delta p_x = -\mu_k F_N\Delta t $ |
| $ y: \Delta p_y = (F_N - mg)\Delta t = 0 $ | $ y: \Delta p_y = (F_N - mg)\Delta t = 0 $ | ||
| + | |||
| + | Write equation of y direction in terms of $F_N$ to sub into x direction equation. | ||
| $ (F_N - mg) \Delta t = 0 $ | $ (F_N - mg) \Delta t = 0 $ | ||
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| Combining these two equations and substituting in mg for $F_N$ and writing $ p_x = \Delta(mv_x) $, we get the following equation: | Combining these two equations and substituting in mg for $F_N$ and writing $ p_x = \Delta(mv_x) $, we get the following equation: | ||
| - | $ \Delta(mv_x) = -mg\Delta t $ | + | $ \Delta(mv_x) = - \mu_k mg\Delta t $ |
| Cancel the masses | Cancel the masses | ||
| - | $ \Delta(v_x) = - g\Delta t $ | + | $ \Delta(v_x) = - \mu_k g\Delta t $ |
| Rearrange to solve for $\Delta t$ and sub in 0 - $v_{xi}$ for $ \Delta(v_x)$ | 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}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k g} $ |
| Fill in values for variables and solve for $\Delta t$ | Fill in values for variables and solve for $\Delta t$ | ||
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| $ \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 $ | ||