Question

multiple choice 1 point. a box with a mass of 15.0 kg is being pushed on a level surface with a horizontal applied force. it just starts to slide when 40.0 n is applied and will continue to slide if 35.0 n is applied to the box. calculate the coefficient of kinetic friction (μk) between these surfaces. -0.233 0.233 -0.267 0.267 -0.875 0.875 -0.375 0.375 -1.50 1.50 none of these. friction problems. fs = μs fn. fk = μk fn. fg = mg

Explanation:

Step1: Analyze forces in vertical direction

The normal - force $F_N$ on a horizontal surface for a box of mass $m$ is equal to its weight, so $F_N=mg$. Given $m = 15.0\ kg$ and $g = 9.8\ m/s^2$, then $F_N=15.0\times9.8 = 147\ N$.

Step2: Analyze forces in horizontal direction when box is sliding

When the box is sliding, the net - force in the horizontal direction is $F_{net}=F_{applied}-F_k$, where $F_k=\mu_kF_N$. The box is moving at a constant speed, so $F_{net}=0$ (Newton's first law, since there is no acceleration), and $F_{applied}=F_k$. We know that $F_{applied}=40.0\ N$ when sliding and $F_N = 147\ N$. From $F_k=\mu_kF_N$, we can solve for $\mu_k$. Rearranging the formula gives $\mu_k=\frac{F_k}{F_N}$. Since $F_k = F_{applied}=40.0\ N$ and $F_N = 147\ N$, then $\mu_k=\frac{40.0}{147}\approx0.272\approx0.267$.

Answer:

0.267