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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:sliding_to_a_stop [2014/09/22 04:30] – 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 ==== | ||
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| === 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 | ||
| + | |||
| + | Write equation of y direction in terms of FN to sub into x direction equation. | ||
| (FN−mg)Δt=0 | (FN−mg)Δt=0 | ||
| - | $ F_N \Delta t - mg \Delta t = 0 \, | + | Multiply out |
| - | $ F_N \Delta t = mg \Delta t | + | $ F_N \Delta t - mg \Delta t = 0 $ |
| - | $ F_N = mg \, | + | Make equal to each other |
| + | |||
| + | $ F_N \Delta t = mg \Delta t $ | ||
| + | |||
| + | Cancel | ||
| + | |||
| + | $ F_N = mg $ | ||
| Combining these two equations and substituting in mg for FN and writing px=Δ(mvx), we get the following equation: | Combining these two equations and substituting in mg for FN and writing px=Δ(mvx), we get the following equation: | ||
| - | Δ(mvx)=−mgΔt | + | $ \Delta(mv_x) = - \mu_k mg\Delta t $ |
| Cancel the masses | Cancel the masses | ||
| - | Δ(vx)=−gΔt | + | $ \Delta(v_x) = - \mu_k g\Delta t $ |
| Rearrange to solve for Δt and sub in 0 - vxi for Δ(vx) | Rearrange to solve for Δt and sub in 0 - vxi for Δ(vx) | ||
| - | Δ(t)=0−vxi−g=vxig | + | $ \Delta(t) = \dfrac{0 - v_{xi}}{-\mu_k g} = \dfrac{v_{xi}}{\mu_k g} $ |
| Fill in values for variables and solve for Δt | Fill in values for variables and solve for Δt | ||
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| 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 | ||