Question

known coefficients
$vec{f}_{s}=mu_{s}vec{f}_{n}$
$vec{f}_{k}=mu_{k}vec{f}_{n}$
$vec{f}_{g}=mvec{a}_{g}$
12 multiple - choice 1 point
a box with a mass of 45.0 kg is being pushed on a level surface with a horizontally - applied force. the coefficient of static friction ($mu s$) is 0.500 and the coefficient of kinetic friction ($mu k$) is 0.400.
how much force ($fk$) must be applied to keep the box sliding?
0.400 n
0.500 n
4.50 n
45.0 n

1. n

225 n

1. n

Explanation:

Step1: Calculate normal force

On a level - surface, $F_N = mg$, where $m = 45.0$ kg and $g= 9.8$ m/s². So $F_N=45.0\times9.8 = 441$ N.

Step2: Calculate kinetic - friction force

The formula for kinetic - friction force is $F_k=\mu_kF_N$. Given $\mu_k = 0.400$ and $F_N = 441$ N. Then $F_k=0.400\times441=176.4\approx180$ N.

Answer:

1. N