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• Notes

$\begin{array}{l}{\text { 14-6. When the driver applics the brakes of a light truck }} \\ {\text { traveling } 40 \mathrm{km} / \mathrm{h}, \text { it skids } 3 \mathrm{m} \text { before stopping. How far will }} \\ {\text { the truck skid if it is traveling } 80 \mathrm{km} / \mathrm{h} \text { when the brakes are }} \\ {\text { applicd? }}\end{array}$

$V_{1}=40 km/h \ , V_2=0$

$d=3 m$

$v'_{1}=80 km/h \ , v_{2}^{'}=0$

$v_{1}=40 km/ h=\frac{40({10}^{3})}{3600}=11.11 m/s$

$v_{1}^{\prime}=80km/ h=\frac{80\left(10^{3}\right)}{3600}=22.22 \mathrm{m} / \mathrm{s}$

$T_{1}+\Sigma U_{1 \rightarrow 2}=T_{2}$

$\frac{1}{2} m v_{1}^{2}+\Sigma u_{1\rightarrow 2}=\frac{1}{2} m v_{2}^{2}$

$\frac{1}{2}(11. 11)^{2}-\mu _k g{(3)}=\frac{1}{2}$

$=\mu_{k} g=20. 576$

$\frac{1}{2} m(22.22)^{2}-\left(20.576\right) m(d)=\frac{1}{2}m(0)$

$∴ d=12 \mathrm{m}$

$\begin{array}{l}{\text { 14-10. The force } \mathbf{f} \text { acting in a comstant direction on the }} \\ {20 \text { . } \mathrm{kg} \text { block. has a masnituale which varies with the position } s} \\ {\text { of the biock. Determes } 15 \mathrm{m} / \mathrm{s} \text { . When } \mathrm{m} \text { the berese }} \\ {\text { incrims to the right at } v=6 \mathrm{m} / \mathrm{s} \text { . The cossorficions of kingtic }} \\ {\text { friction between the block and surface is } \mu_{k}=0.3 \text { . }}\end{array}$

$m=20 kg$

$f=50S^{\frac{1}{2}}$

$v_{2}=15 \mathrm{m} / \mathrm{s}$

$@ S=0 \rightarrow V_{1}=6 m / S$

$\mu_{k}=0.3$

$+ \uparrow \Sigma Fy=0 \rightarrow N-w=0 \rightarrow N=20(9.81)=196.2 N$

$F_{f}=\mu_{k} N=0.3(196.2)=58.86 \mathrm{N}$

$U_{f}=\int_{0}^{s} F d s=\int_{0}^{s} 50s^{\frac{1}{2}} d s=(+) \frac{100}{3} s^{\frac{3}{2}}$

$U_{F_f}=F_f*S=-58.86 S$

$T_1+|Sigma U_{1 \rightarrow 2}=T_{2} \Rightarrow \frac{1}{2} m v_{1}^{2}+\left(U_{f}+U_{f_{f}}\right)=\frac{1}{2} m v_{2}^{2}$

$\frac{1}{2}(20)\left(6^{2}\right)+\left(\frac{100}{3} s^{\frac{3}{2}}\right)+(-58.865)=\frac{1}{2}(20)(15)^{2}$

$\frac{100}{3}s^{\frac{3}{2}} -58.865-1890=0$

$S \simeq20.5 \mathrm{m}$

$\begin{array}{l}{14-11 . \text { The force of } F=50 \mathrm{N} \text { is applied to the cord when }} \\ {s=2 \mathrm{m} . \text { If the } 6-\mathrm{kg} \text { collar is orginally at rest, determine its }} \\ {\text { velocity at } s=0 . \text { Neglect friction. }}\end{array}$

$F =50 N$

$s_{1}=2 m$

$m=6 k g$

$v_{1}=0$

$S_{2}=0$

$d=\sqrt{s^{2}+1.5^{2}}=\sqrt{2^{2}+1 \cdot 5^{2}}=2. 5 m$

$s^{\prime}=2.5- 1.5=1\mathrm{m}$

$U_{F}=F \cdot S^{\prime}=50 * 1=50 J$

$T_{1}+\Sigma U_{1 \rightarrow2}=T_{2} \Rightarrow \frac{1}{2} mV_1+U_{F}=\frac{1}{2} m v_{2}$

$\frac{1}{2} 6(0)+50=\frac{1}{2} 6\left(v_{2}^{2}\right)$

$. v=4.082 \mathrm{m} / \mathrm{s} \simeq 4.08 \mathrm{m} / \mathrm{s}$