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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:sliding_to_a_stop [2014/09/16 07:51] – 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 ⟨6,0,0⟩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 ⟨6,0,0⟩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 | Mass of metal block = 3 kg | ||
| Line 10: | Line 12: | ||
| The coefficient of friction between floor and block = 0.4 | The coefficient of friction between floor and block = 0.4 | ||
| - | Initial velocity of block ⟨6,0,0⟩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: | ||
| Δ→p=→FnetΔt | Δ→p=→FnetΔt | ||
| Line 22: | Line 34: | ||
| === Solution === | === Solution === | ||
| - | x:Δpx=−FNΔt | + | $ x: \Delta p_x = -\mu_k F_N\Delta t $ |
| y:Δpy=(FN−mg)Δt=0 | y:Δpy=(FN−mg)Δt=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 $ |
| - | Δ(vx)=−gΔt | + | Multiply out |
| - | Δ(t)=0−vxi−g=vxig | + | FNΔt−mgΔt=0 |
| + | |||
| + | Make equal to each other | ||
| + | |||
| + | FNΔt=mgΔt | ||
| + | |||
| + | Cancel Δt | ||
| + | |||
| + | FN=mg | ||
| + | |||
| + | Combining these two equations and substituting in mg for FN and writing px=Δ(mvx), we get the following equation: | ||
| + | |||
| + | Δ(mvx)=−μkmgΔt | ||
| + | |||
| + | Cancel the masses | ||
| + | |||
| + | Δ(vx)=−μkgΔt | ||
| + | |||
| + | Rearrange to solve for Δt and sub in 0 - vxi for Δ(vx) | ||
| + | |||
| + | $ \Delta(t) = \dfrac{0 - v_{xi}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k g} $ | ||
| + | |||
| + | Fill in values for variables and solve for $\Delta t$ | ||
| Δ(t)=6m/s0.4(9.8N/kg)=1.53s | Δ(t)=6m/s0.4(9.8N/kg)=1.53s | ||
| - | Since the net force was constant, vx,avg=(vxi+vxf)/2, so | + | Since the net force was constant |
| Δx/Δt=((6+0)/2)m/s=3m/s | Δx/Δt=((6+0)/2)m/s=3m/s | ||
| + | |||
| + | Sub in for Δt and solve for Δx | ||
| Δx=(3m/s)(1.53s)=4.5m | Δx=(3m/s)(1.53s)=4.5m | ||