Question

chart look up
$vec{f}_s=mu_svec{f}_n$
$vec{f}_k=mu_kvec{f}_n$
$vec{f}_g = mvec{a}_g$
13 fill in the blank 2 points
a rubber box with a mass of 31.0 kg is being pushed on a level dry concrete surface with a horizontally applied force.
how much force (fs) must be applied to start the box sliding? type your answer.
(be sure your answer has the appropriate number of significant figures) (include unit: n, kg, m/s^2) (put a space in between number and unit; example: 10 m/s^2)

Explanation:

Step1: Determine the normal force

The normal force $F_N$ on a level surface is equal to the weight of the object. Using $F_g = m\vec{a}_g$, and since $a_g = g= 9.8\ m/s^2$ and $F_N=F_g$, we have $F_N=mg$. Given $m = 31.0\ kg$, then $F_N=31.0\ kg\times9.8\ m/s^2 = 303.8\ N$.

Step2: Find the coefficient of static - friction

From the chart, for rubber on dry concrete, the coefficient of static - friction $\mu_s = 1.0$.

Step3: Calculate the static - friction force

The maximum static - friction force $F_s=\mu_sF_N$. Substituting $\mu_s = 1.0$ and $F_N = 303.8\ N$ into the formula, we get $F_s=1.0\times303.8\ N=303.8\ N\approx304\ N$.

Answer:

304 N