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183_notes:examples:sliding_to_a_stop [2014/09/16 07:40] – 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 ==== | ||
+ | |||
+ | Block is metal. | ||
+ | |||
+ | Mass of metal block = 3 kg | ||
+ | |||
+ | The coefficient of friction between floor and block = 0.4 | ||
+ | |||
+ | 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: | ||
+ | |||
+ | $\Delta \vec{p} = \vec{F}_{net} \Delta t$ | ||
=== 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 $ | ||
- | Combining these two equations and writing | + | Write equation of y direction in terms of $F_N$ to sub into x direction equation. |
- | $ \Delta(mv_x) = -mg\Delta t $ | + | $ (F_N - mg) \Delta t = 0 $ |
- | $ \Delta(v_x) = - g\Delta t $ | + | Multiply out |
- | $ \Delta(t) = \dfrac{0 - v_{xi}}{-g} = \dfrac{v_{xi}}{g} $ | + | $ 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 and substituting in mg for $F_N$ and writing $ p_x = \Delta(mv_x) $, we get the following equation: | ||
+ | |||
+ | $ \Delta(mv_x) = - \mu_k mg\Delta t $ | ||
+ | |||
+ | Cancel the masses | ||
+ | |||
+ | $ \Delta(v_x) = - \mu_k 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}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k 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 we can say the average velocity can be described as: $v_{x,avg} = (v_{xi} + v_{xf})/2$, so | ||
+ | |||
+ | $ \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 $ |