Question

if a 3 kg block rests on a rough horizontal table and experiences a 2 n force of static friction, then what is the coefficient of static friction between the block and the table? use g = 9.81 m/s² and round to the nearest hundredths.

Explanation:

Step1: Calculate the normal force

The normal force $N$ on a horizontal surface for a block of mass $m$ is given by $N = mg$. Here, $m = 3\ kg$ and $g=9.81\ m/s^{2}$, so $N=3\times9.81 = 29.43\ N$.

Step2: Use the static - friction formula

The formula for static friction is $F_f=\mu_sN$, where $F_f$ is the force of static friction, $\mu_s$ is the coefficient of static friction, and $N$ is the normal force. We know $F_f = 2\ N$ and $N = 29.43\ N$. Rearranging for $\mu_s$, we get $\mu_s=\frac{F_f}{N}$. Substituting the values, $\mu_s=\frac{2}{29.43}\approx0.07$.

Answer:

$0.07$