Question

1. esteban pushes a 47 kg box across a concrete parking lot. if the coefficient of kinetic friction is 0.51, then what is the force of friction on the box?

23.97 n
5.00 n
234.91 n
460.60 n

Explanation:

Step1: Calculate the normal force

The normal force $N$ on a flat - surface is equal to the weight of the object. Using the formula $N = mg$, where $m = 47\ kg$ and $g=9.8\ m/s^{2}$. So $N=47\times9.8 = 460.6\ N$.

Step2: Calculate the force of kinetic friction

The formula for the force of kinetic friction is $F_f=\mu_kN$. Given $\mu_k = 0.51$. Then $F_f=0.51\times460.6$.
$F_f = 0.51\times460.6=234.906\approx234.91\ N$.

Answer:

234.91 N