a car with a mass of 1.8×10³ kg is parked on a hill in san francisco. suppose the hill makes a 15.0° incline with the horizontal. if the frictional force required to keep the car from sliding down the hill is 1.25×10⁴ n, what is the coefficient of static friction between the pavement and the cars tires?

Question

Accepted Answer

# Explanation:
## Step1: Calculate the normal force
The normal force $N$ on an inclined plane is $N = mg\cos\theta$, where $m = 1.8\times10^{3}\text{ kg}$, $g = 9.8\text{ m/s}^2$, and $\theta=15.0^{\circ}$.
$$
\begin{align*}
N&=(1.8\times 10^{3})\times9.8\times\cos(15^{\circ})\
&=(1.8\times 10^{3})\times9.8\times0.9659\
&=1.8\times9.8\times0.9659\times 10^{3}\
&=17.64\times0.9659\times 10^{3}\
&\approx1.704\times 10^{4}\text{ N}
\end{align*}
$$
## Step2: Use the formula for static frictional force
The formula for static frictional force is $f_s=\mu_sN$, where $f_s = 1.25\times 10^{4}\text{ N}$ and $N\approx1.704\times 10^{4}\text{ N}$. Solving for $\mu_s$, we get $\mu_s=\frac{f_s}{N}$.
$$
\begin{align*}
\mu_s&=\frac{1.25\times 10^{4}}{1.704\times 10^{4}}\
&=\frac{1.25}{1.704}\
&\approx0.734
\end{align*}
$$

# Answer:
The coefficient of static friction $\mu_s\approx0.734$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the normal force

Step2: Use the formula for static frictional force

Answer: