Question

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 (μs) is 0.500 and the coefficient of kinetic friction (μk) is 0.400. how much force (fs) must be applied to start the box sliding? 0.400 n 0.500 n 4.50 n 45.0 n 180. n 225 n 450. n known coefficients (vec{f}_s=mu_svec{f}_n) (vec{f}_k = mu_kvec{f}_n) (vec{f}_g=mg)

Explanation:

Step1: Calculate the normal force

The box is on a level surface, so the normal force $F_N$ equals the weight of the box. Using $F_g = mg$, where $m = 45.0\ kg$ and $g= 9.8\ m/s^2$, we have $F_N=F_g=mg = 45.0\times9.8\ N=441\ N$.

Step2: Calculate the static - friction force

The formula for the maximum static - friction force is $F_s=\mu_sF_N$. Given $\mu_s = 0.500$, then $F_s=0.500\times441\ N = 220.5\ N\approx225\ N$. This is the minimum force required to start the box sliding.

Answer:

225 N