Question

question 17
a man puts a box into the trunk of his car. later, he drives around a corner that has a radius of 60 m. the speed of the car around the curve is 21 m/s and the box remains stationary relative to the floor of the trunk. determine the minimum coefficient of static friction between the box and the floor.
there is not enough information
0.75
0.036
0.24
0.45

Explanation:

Step1: Identify centripetal - force formula

The centripetal force $F_c$ acting on the box is provided by the frictional force $F_f$. The centripetal - force formula is $F_c=\frac{mv^{2}}{r}$, and the frictional - force formula is $F_f = \mu_sN$, where $N = mg$ (on a horizontal surface). So, $\frac{mv^{2}}{r}=\mu_smg$.

Step2: Solve for the coefficient of static friction

Cancel out the mass $m$ from both sides of the equation $\frac{mv^{2}}{r}=\mu_smg$. We get $\mu_s=\frac{v^{2}}{gr}$. Given $v = 21$ m/s, $r = 60$ m, and $g=9.8$ m/s². Substitute the values: $\mu_s=\frac{21^{2}}{9.8\times60}$.

Step3: Calculate the value of $\mu_s$

First, calculate $21^{2}=441$, then $9.8\times60 = 588$. So, $\mu_s=\frac{441}{588}=0.75$.

Answer:

0.75