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$$
\begin{array}{l}{\text { 13. G. The } 10-1 b \text { block has a speed of } 4 \mathrm{ft} / \mathrm{s} \text { when the force }} \\ {\text { of } F=\left(8 t^{2}\right) \text { ib is applied. Determine the velocity of the }} \\ {\text { block when } t=2 \text { s. The coefficient of kinetic friction at the }} \\ {\text { surface is } \mu_{k}=0.2 \text { . }}\end{array}
$$

$$
w=10l b
$$

$$
v=4 \mathrm{ft} / \mathrm{s}
$$

$$
F=8 t^{2} l b
$$

$$
t=2s
$$

$$
\mu=0.2
$$

$$
 + \uparrow \Sigma F y=m a y \rightarrow N-10=\frac{10}{32.2} ay
$$

$$
N=10lb
$$

$$
\stackrel{+}{\rightarrow}\ \Sigma F_{x}=m a_{x} \rightarrow 8 t^{2}-o \cdot 2(10)=\frac{10}{32.2} * a
$$

\(∴ a=3.32\left(8 t^{2}-2\right) f t / s^{2} \)

$$
a=\frac{d v}{d t} \rightarrow d v=a d t
$$

@ $$
t=0 \rightarrow \ v=4ft
$$

$$
\int_{4ft}^v d v=\int_{0}^{ t} 3.22\left(8 t^{2}-2\right) d t
$$

$$
v-4=3.22\left(\frac{8}{3} t^{3}-2 t\right)
$$

$$
=8.586 t^{3}-6.44 t+4 \quad @ t=2s
$$

$$
V=59.8 \mathrm{ft} / \mathrm{s}
$$

$$
\begin{array}{l}{\text { 13-10. The conveyor belt is designed to transport packages }} \\ {\text { of various weights. Each } 10-\mathrm{kg} \text { package has a coefficient of }} \\ {\text { kinetic friction } \mu_{k}=0.15 . \text { If the speed of the conveyor is }} \\ {5 \mathrm{m} / \mathrm{s} \text { , and then it suddenly stops, determine the distance the }} \\ {\text { package will slide on the belt before coming to rest. }}\end{array}
$$

$$
m=10 kg
$$

$$
\mu_{k}=0.15
$$

$$
v_{0}=5 \mathrm{m} / \mathrm{s}
$$

$$
V=0
$$

$$
+\uparrow \Sigma F y=m a y \Rightarrow N-W=may
$$

\(∴ N=W=98.1 \mathrm{N} \)

\( \stackrel{+}{\rightarrow} \Sigma F x=\max \Rightarrow 0.15*98.1=10*a \Rightarrow ∴ a=1.4715 m/s^2 \)

$$
v^{2}=v_{0}^{2}+2 a_{c}\left(s-s_{0}\right)
$$

$$
0=(5)^{2}+2(-1.4715)(s-0)
$$

\(∴ s=8.49 m \)

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