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$$
\begin{array}{l}{\text { 15-13. The 2.5-Mg van is traveling with a speed of } 100 \mathrm{km} / \mathrm{h}} \\ {\text { when the brakes are applied and all four wheels lock. If the }} \\ {\text { speed decreases to 40 } \mathrm{km} / \mathrm{h} \text { in 5 s, determine the coefficient }} \\ {\text { of kinetic friction between the tires and the road. }}\end{array}
$$

$$
m=2.5 \mathrm{Mg}=2.5 \times 10^{6} \mathrm{g}
$$

$$
=2500 \mathrm{kg}
$$

$$
v_{1}=100 \mathrm{km} / \mathrm{h}
$$

$$
v_{2}=40 k{m} / h
$$

$$
t_2=5s \ , t_1=0
$$

$$
\mu_{k}=? ?
$$

$$
v_{1}=100 * \frac{1000}{3600}=27.78m/ s, \  v_{2}=40 * \frac{1000}{3600}=11.11 m/ \mathrm{s}
$$

\( (+ \uparrow y) ∴ m (v_{1})_{y}+\Sigma \int_{t_1}^{t_{2}}F_{y} d t= m (v_{2})_ y \)

$$
2500(0)+[N(5-0)- 2500 (9.81)(5-1)]=2500 (0)
$$

$$
N=24525 \mathrm{N}
$$

\( (\stackrel{+}{\leftarrow} x)∴ m\left(v_{1}\right)_{x}+\Sigma F_{x}\left(t_{2}-t_1\right)=m\left(v_{2}\right)_{x} \)

$$
2500(27 . 78) +[-\mu _k(24525)(5-0)]=2500(11 . 11)
$$

$$
\mu_{k}=0.340
$$

$$
\begin{array}{l}{\text { 15-23. The motor pulls on the cable at } A \text { with a force }} \\ {F=\left(30+t^{2}\right) \text { lb, where } t \text { is in seconds. If the } 34-1 b \text { crate is }} \\ {\text { originally on the ground at } t=0 \text { , determine its speed in } t=4 \text { s. }} \\ {\text { Neglect the mass of the cable and pulleys. Hint: First find }} \\ {\text { the time needed to begin lifting the crate. }}\end{array}
$$

$$
F=\left(30+t^{2}\right) lb
$$

$$
w=34lb
$$

$$
t^{\prime}=0
$$

$$
t_{2}=4s
$$

$$
F=w \Rightarrow 30+t^{2}=34 \Rightarrow t^{2}=4 \Rightarrow t=2
$$

$$
(+ \uparrow) m v_{1}+\Sigma \int F d t=m v_{2}
$$

$$
0+\left[\int_{t_{1}}^{t_2} f d t-w\left(t_{2}-t_1\right)\right]=m v_{2}
$$

$$
0+\int_{2}^{4}\left(30+t^{2}\right) d t-34(4-2)=\frac{34}{32.2} v_{2}
$$

$$
\left[30 t+\frac{1}{3} t^{3}\right]_{2}^{4}-68=\frac{34}{32.2} \mathrm{V}_{2}
$$

$$
v_{2}=10.1 ft/s
$$

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