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

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

$$
d=3 m
$$

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

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

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

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

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

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

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

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

\(∴ d=12 \mathrm{m} \)

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

$$
m=20 kg
$$

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

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

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

$$
\mu_{k}=0.3
$$

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

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

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

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

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

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

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

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

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

$$
F =50 N
$$

$$
s_{1}=2 m
$$

$$
m=6 k g
$$

$$
v_{1}=0
$$

$$
S_{2}=0
$$

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

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

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

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

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

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

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