Question

if a 35 n block is resting on a steel table with a coefficient of static friction $mu_{s}=0.40$, then what minimum force is required to move the block.
___ n
do not include units in your answer.

Explanation:

Step1: Recall static - friction formula

The formula for the maximum static - friction force is $F_f=\mu_sN$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

Step2: Identify normal force

Since the block is on a horizontal table and is not accelerating vertically, the normal force $N$ equals the weight of the block. Given that the weight of the block $W = 35\ N$, so $N = 35\ N$.

Step3: Calculate the static - friction force

Substitute $\mu_s=0.40$ and $N = 35\ N$ into the formula $F_f=\mu_sN$. We get $F_f=0.40\times35$.
$F_f = 14$

Answer:

14