Question

1. a car has a mass of 1315 kg. the coefficient of kinetic friction between the cars tires and the road is 0.83. if the car was to skid to a stop, what would the force of friction be while the car was sliding?

$f_f = 1090n$

$f_f = 10,700n$

$f_f = 1580n$

$f_f = 15,500n$

Explanation:

Step1: Calculate the normal force

The normal force \(F_N\) on a horizontal surface is equal to the weight of the object. Using the formula \(F_N = mg\), where \(m = 1315\space kg\) and \(g= 9.8\space m/s^{2}\).
\(F_N=1315\times9.8 = 12887\space N\)

Step2: Calculate the force of kinetic friction

The formula for the force of kinetic friction is \(F_f=\mu_kF_N\). Given \(\mu_k = 0.83\) and \(F_N = 12887\space N\)
\(F_f=0.83\times12887\approx10700\space N\)

Answer:

\(F_f = 10,700\space N\) (the second option)