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? clear all 5.00 n 234.91 n 460.60 n 23.97 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 $m = 4.7$ kg and $g=9.8$ m/s². So $N=4.7\times9.8 = 46.06$ N.

Step2: Calculate the frictional force

The formula for the kinetic frictional force is $F_f=\mu_kN$. Given $\mu_k = 0.51$ and $N = 46.06$ N. Then $F_f=0.51\times46.06=23.4906\approx23.97$ N.

Answer:

23.97 N