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? 5.00 n 234.91 n 23.97 n 460.60 n clear all

Explanation:

Step1: Calculate the normal force

The normal force $F_N$ on a horizontal surface for an object of mass $m$ is given by $F_N = mg$, where $g = 9.8\ m/s^2$ and $m = 47\ kg$. So $F_N=47\times9.8 = 460.6\ N$.

Step2: Calculate the force of kinetic friction

The formula for the force of kinetic friction $F_f$ is $F_f=\mu_kF_N$, where $\mu_k = 0.51$ and $F_N = 460.6\ N$. So $F_f=0.51\times460.6=234.906\approx234.91\ N$.

Answer:

234.91 N