Question

1. tipler pse5 05 38 10 pts possible

the coefficient of static friction between the bed of a truck and a box resting on it is 0.5. the truck is traveling at 130 km/h along a horizontal road.
what is the shortest distance in which the truck can stop if the box is not to slide? the acceleration of gravity is 9.81 m/s².
answer in units of m. answer in units of m.

Explanation:

Step1: Find the maximum acceleration of the truck without the box sliding

The maximum static - friction force $F_f=\mu_sN$, where $N = mg$ (since the box is on a horizontal surface). According to Newton's second - law $F = ma$, and the maximum force causing acceleration of the box (without sliding) is the static - friction force. So $ma=\mu_smg$, and the maximum acceleration $a=\mu_sg$. Given $\mu_s = 0.5$ and $g = 9.81\ m/s^2$, then $a=0.5\times9.81\ m/s^2 = 4.905\ m/s^2$.

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 $v = 0\ m/s$.

Step3: Use the kinematic equation $v^{2}-v_{0}^{2}=2ax$

We want to find the distance $x$. Rearranging the kinematic equation $v^{2}-v_{0}^{2}=2ax$ for $x$, we get $x=\frac{v^{2}-v_{0}^{2}}{2a}$. Substituting $v = 0$, $v_0\approx36.11\ m/s$, and $a=- 4.905\ m/s^2$ (negative because it is decelerating), we have $x=\frac{0-(36.11)^2}{2\times(-4.905)}$.
\[

$$\begin{align*} x&=\frac{- 1303.9321}{-9.81}\\ x&=133\ m \end{align*}$$

\]

Answer:

$133$