Vraag 6

voorbeelden van een antwoord:

methode 1

Ten gevolge van de schuifwrijvingskracht neemt de kinetische energie van het rotsblok af. Er geldt$\Delta E_{\mathrm{k}}=\sum Wmet$\Delta E_{\mathrm{k}}=(-) \frac{1}{2} m v_{\text {begin }}^{2}en$\sum W=W_{\mathrm{w}}=(-) F_{\mathrm{w}} s. De schuifwrijvingskracht$F_{\mathrm{w}}is gelijk aan$0{,}25 \cdot F_{\mathrm{N}}=0{,}25 \cdot F_{\mathrm{z}}=0{,}25 \cdot m g.

Hieruit volgt$s=\frac{\Delta E_{k}}{F_{w}}=\frac{\frac{1}{2} m v_{\text {begin }}^{2}}{0{,}25 \cdot m g}=\frac{2 v_{\text {begin }}^{2}}{g}=\frac{2 \cdot 14^{2}}{5{,}7 \cdot 10^{-3}}=6{,}9 \cdot 10^{4} \mathrm{~m}.

methode 2

Omdat de schuifwrijvingskracht constant is, is de beweging eenparig vertraagd. Uit F_{res}=maF_{res}=mF_{res}=F_{res}F_{re}F_{r}F en F_{w}=0{,}25\cdot mgF_{w}=0{,}25\cdot mF_{w}=0{,}25\cdotF_{w}=0{,}25F_{w}=0{,}2F_{w}=0{,}F_{w}=0F_{w}=F_{w}Fvolgt a=0{,}25ga=0{,}25a=0{,}2a=0{,}a=0a=a. Met a_{gem}=\frac{\Delta v}{\Delta t}a_{gem}=\frac{\Delta v}{\Delta}a_{gem}=\frac{\Delta v}{\placeholder{}}a_{gem}=\Delta va_{gem}=\Deltaa_{gem}=a_{gem}a_{ge}a_{g}avolgt hieruit voor de remtijd: begin eind\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\cdot10^3s\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\cdot10^3\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\cdot10\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\cdot1\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\cdot\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}82\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}8\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9{,}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=9\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}=\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-3}}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10^{-}}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot10}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot1}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7\cdot}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}7}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5{,}}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot5}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25\cdot}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}25}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}2}\Delta t=\frac{\Delta v}{a}=\frac{14}{0{,}}\Delta t=\frac{\Delta v}{a}=\frac{14}{0}\Delta t=\frac{\Delta v}{a}=\frac{14}{\placeholder{}}\Delta t=\frac{\Delta v}{a}=14\Delta t=\frac{\Delta v}{a}=1\Delta t=\frac{\Delta v}{a}=\Delta t=\frac{\Delta v}{a}\Delta t=\frac{\Delta v}{\placeholder{}}\Delta t=\Delta v\Delta t=\Delta\Delta t=\Delta t\Delta .

De gemiddelde snelheid is v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}0ms^{-1}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}0ms^{-}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}0msv_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}0mv_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}0v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7{,}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=7v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}=v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+0}{2}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14+}{2}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14}{2}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=\frac{14}{\placeholder{}}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=14v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=1v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}=v_{gem}=\frac{v_{begin}+v_{eind}_{}}{2}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{}v_{gem}=\frac{v_{begin}+v_{eind}_{}}{n}v_{gem}=\frac{v_{begin}+v_{}}{n}v_{gem}=\frac{v_{begin}+v_{e}}{n}v_{gem}=\frac{v_{begin}+v_{e\in}}{n}v_{gem}=\frac{v_{begin}+v_{ei}}{n}v_{gem}=\frac{v_{begin}+v_{e}}{n}v_{gem}=\frac{v_{begin}+v}{n}v_{gem}=\frac{v_{begin}+}{n}v_{gem}=\frac{v_{begin}}{n}v_{gem}=\frac{v_{begin+}}{n}v_{gem}=\frac{v_{begin+v}}{n}v_{gem}=\frac{v_{begin+}}{n}v_{gem}=\frac{v_{begin}}{n}v_{gem}=\frac{v_{begin=}}{n}v_{gem}=\frac{v_{begin}}{n}v_{gem}=\frac{v_{begin+}}{n}v_{gem}=\frac{v_{begin}}{n}v_{gem}=\frac{v_{}}{n}v_{gem}=\frac{v_{b}}{n}v_{gem}=\frac{v_{be}}{n}v_{gem}=\frac{v_{beg}}{n}v_{gem}=\frac{v_{beg\in}}{n}v_{gem}=\frac{v_{begi}}{n}v_{gem}=\frac{v_{beg}}{n}v_{gem}=\frac{v_{be}}{n}v_{gem}=\frac{v_{b}}{n}v_{gem}=\frac{v}{n}v_{gem}=\frac{v}{\placeholder{}}v_{gem}=vv_{gem}=v_{gem}v_{ge}v_{g}v , dus de remafstand is \Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}9\cdot10^4\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}9\cdot10\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}9\cdot1\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}9\cdot\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}9\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6{,}\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=6\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3=\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10^3\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot10\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot1\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\cdot\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}82\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}8\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9{,}\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot9\Delta x=v_{gem}\cdot\Delta t=7{,}0\cdot\Delta x=v_{gem}\cdot\Delta t=7{,}0\Delta x=v_{gem}\cdot\Delta t=7{,}\Delta x=v_{gem}\cdot\Delta t=7\Delta x=v_{gem}\cdot\Delta t=\Delta x=v_{gem}\cdot\Delta t\Delta x=v_{gem}\cdot\Delta\Delta x=v_{gem}\cdot\Delta x=v_{gem}\Delta x=v_{ge}\Delta x=v_{g}\Delta x=v\Delta x=\Delta x\Delta