Question

chart look up
$vec{f}_s=mu_svec{f}_n$
$vec{f}_k=mu_kvec{f}_n$
$vec{f}_g = mvec{a}_g$
14 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 (fk) must be applied to keep the box sliding at a constant velocity?
type your answer...
(be sure your answer has the appropriate number of significant figures) (include unit: n, kg, m/s^2) (put a sp between number and unit, example: -10 m/s^2)

Explanation:

Step1: Calculate the normal force

The normal force $\vec{F}_N$ on a level surface is equal to the gravitational force $\vec{F}_g$. Using $F_N = F_g=mg$, where $m = 31.0$ kg and $g = 9.8$ m/s². So $F_N=31.0\times9.8 = 303.8$ N.

Step2: Determine the coefficient of kinetic friction

From the table, for rubber on dry - concrete, $\mu_k = 0.8$.

Step3: Calculate the kinetic friction force

Using the formula $F_k=\mu_kF_N$, substitute $F_N = 303.8$ N and $\mu_k = 0.8$. So $F_k=0.8\times303.8 = 243.04$ N. Rounding to three significant figures, $F_k = 243$ N.

Answer:

243 N