Question

a 1.00×10³ - n crate pushed across a level floor at constant speed by force f of 3.00×10²n at 20.0° angle below horizontal. what is the coefficient of kinetic friction between crate and floor?

Explanation:

Step1: Analyze vertical - force equilibrium

Since the crate is not moving vertically ($a_y = 0$), the sum of vertical forces $\sum F_y=0$. The vertical forces are the normal force $F_N$, the weight of the crate $F_g = 1.00\times10^{3}\ N$, and the vertical component of the applied force $F\sin20^{\circ}$ (down - ward). So, $F_N=F_g + F\sin20^{\circ}$.
$F_N=1.00\times 10^{3}+3.00\times 10^{2}\sin20^{\circ}$
$F_N = 1000+300\times0.3420 = 1000 + 102.6=1102.6\ N$

Step2: Analyze horizontal - force equilibrium

Since the crate is moving at a constant speed ($a_x = 0$), the sum of horizontal forces $\sum F_x = 0$. The horizontal forces are the horizontal component of the applied force $F\cos20^{\circ}$ and the kinetic - friction force $F_k$. So, $F_k=F\cos20^{\circ}$.
$F_k = 3.00\times 10^{2}\cos20^{\circ}=300\times0.9397 = 281.91\ N$

Step3: Calculate the coefficient of kinetic friction

The formula for the kinetic - friction force is $F_k=\mu_kF_N$. Then, $\mu_k=\frac{F_k}{F_N}$.
$\mu_k=\frac{281.91}{1102.6}\approx0.256$

Answer:

$0.256$