Question

if a 20 n block experiences a force of kinetic friction of 8.0 n from the surface of a table, then what is the coefficient of kinetic friction ($mu_{k}$) between the block and the table?

Explanation:

Step1: Recall friction formula

The formula for kinetic - friction is $F_k=\mu_kF_N$, where $F_k$ is the force of kinetic friction, $\mu_k$ is the coefficient of kinetic friction, and $F_N$ is the normal force.

Step2: Identify the normal force and kinetic - friction force

The weight of the block is $20\ N$. On a horizontal surface, the normal force $F_N$ equals the weight of the block, so $F_N = 20\ N$, and the force of kinetic friction $F_k=8.0\ N$.

Step3: Solve for the coefficient of kinetic friction

Rearrange the formula $F_k=\mu_kF_N$ to $\mu_k=\frac{F_k}{F_N}$. Substitute $F_k = 8.0\ N$ and $F_N=20\ N$ into the formula: $\mu_k=\frac{8.0}{20}=0.4$.

Answer:

$0.4$