Question

1. a 1.00×10³ - n crate is being pushed across a level floor at a constant speed by a force (vec{f}) of 3.00×10² n at an angle of 20.0° below the horizontal, as shown in figure p4.23a. (a) what is the coefficient of kinetic friction between the crate and the floor? (b) if the 3.00×10² - n force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in figure p4.23b, what will be the acceleration of the crate? assume that the coefficient of friction is the same as that found in part (a).

Explanation:

Step1: Analyze forces in the first - case (constant speed)

In the x - direction, $F\cos20^{\circ}-\mu_kN = 0$ (since $a = 0$). In the y - direction, $N - mg - F\sin20^{\circ}=0$, so $N=mg + F\sin20^{\circ}$. Given $mg = 1.00\times10^{3}\ N$ and $F = 3.00\times10^{2}\ N$. First, find $N$:
$N=1.00\times10^{3}+3.00\times10^{2}\sin20^{\circ}=1.00\times10^{3}+3.00\times10^{2}\times0.342 = 1.00\times10^{3}+102=1102\ N$. Then, from $F\cos20^{\circ}-\mu_kN = 0$, we can solve for $\mu_k$: $\mu_k=\frac{F\cos20^{\circ}}{N}$. Substitute $F = 3.00\times10^{2}\ N$, $\cos20^{\circ}=0.9397$ and $N = 1102\ N$ into the formula:
$\mu_k=\frac{3.00\times10^{2}\times0.9397}{1102}=\frac{281.91}{1102}\approx0.256$.

Step2: Analyze forces in the second - case

In the x - direction, $F\cos20^{\circ}-\mu_kN=ma$. In the y - direction, $N + F\sin20^{\circ}-mg = 0$, so $N=mg - F\sin20^{\circ}=1.00\times10^{3}-3.00\times10^{2}\times0.342=1.00\times10^{3}-102 = 898\ N$. We know $\mu_k\approx0.256$, $F = 3.00\times10^{2}\ N$, $\cos20^{\circ}=0.9397$ and $m=\frac{mg}{g}=\frac{1.00\times10^{3}}{9.8}\ kg$.
First, calculate the net force in the x - direction: $F_{net,x}=F\cos20^{\circ}-\mu_kN=3.00\times10^{2}\times0.9397-0.256\times898=281.91 - 230.89=51.02\ N$.
Then, since $m=\frac{1.00\times10^{3}}{9.8}\ kg\approx102.04\ kg$, from $F_{net,x}=ma$, we can find $a$: $a=\frac{F_{net,x}}{m}=\frac{51.02}{102.04}=0.5\ m/s^{2}$.

Answer:

(a) $\mu_k\approx0.256$; (b) $a = 0.5\ m/s^{2}$