9. tipler pse5 05 38 10 pts possible\nthe coefficient of static friction between the bed of a truck and a…

Answer:

135.29

Explanation:

Step1: Find the maximum acceleration without sliding

The maximum static - friction force $F_f=\mu_sN$, where $N = mg$ (since the normal force equals the weight on a horizontal surface). According to Newton's second - law $F = ma$, and $F_f=ma_{max}$. So $ma_{max}=\mu_smg$, and $a_{max}=\mu_sg$. Given $\mu_s = 0.5$ and $g = 9.81\ m/s^2$, then $a_{max}=0.5\times9.81\ m/s^2 = 4.905\ m/s^2$.

Step2: Convert the initial velocity to SI units

The initial velocity $v_0=130\ km/h$. To convert it to $m/s$, we use the conversion factor: $1\ km = 1000\ m$ and $1\ h=3600\ s$. So $v_0 = 130\times\frac{1000}{3600}\ m/s=\frac{1300}{36}\ m/s\approx36.11\ m/s$. The final velocity $v = 0$.

Step3: Use the kinematic equation $v^{2}-v_{0}^{2}=2a\Delta x$

We want to find the displacement $\Delta x$. Rearranging the kinematic equation $v^{2}-v_{0}^{2}=2a\Delta x$ for $\Delta x$, we get $\Delta x=\frac{v^{2}-v_{0}^{2}}{2a}$. Since $v = 0$ and $a=-a_{max}$ (negative because it's decelerating), then $\Delta x=\frac{0 - v_{0}^{2}}{-2a_{max}}$. Substituting $v_0\approx36.11\ m/s$ and $a_{max}=4.905\ m/s^2$ into the formula: $\Delta x=\frac{36.11^{2}}{2\times4.905}=\frac{1303.93}{9.81}=135.29\ m$.