Question

1. -/20 points a horizontal force of 150 n is used to push a 48.0 - kg packing crate a distance of 5.15 m on a rough horizontal surface. if the crate moves at constant speed, find each of the following. (a) the work done by the 150 - n force (b) the coefficient of kinetic friction between the crate and the surface resources read it details my notes ask your teacher practice another assignment scoring your best submission for each question part is used for your score. for this assignment, you submit answers by question parts. the number of submissions remaining for each question part only changes if you submit or change the answer.

Explanation:

Step1: Recall work - formula

The work - done formula is $W = Fd\cos\theta$, where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement. Here, the force $F = 150\ N$, the displacement $d=5.15\ m$, and since the force is horizontal and the motion is horizontal, $\theta = 0^{\circ}$ and $\cos\theta=\cos(0^{\circ}) = 1$.
$W=Fd\cos\theta=150\times5.15\times1$

Step2: Calculate the work

$W = 150\times5.15=772.5\ J$

For part (b), since the crate moves at a constant speed, the net force on the crate is zero. The applied force $F = 150\ N$ is equal to the kinetic - friction force $f_k$. The normal force $N$ on the crate on a horizontal surface is equal to its weight, $N = mg$, where $m = 48.0\ kg$ and $g = 9.8\ m/s^{2}$.
$N=mg=48.0\times9.8 = 470.4\ N$
The kinetic - friction force $f_k=\mu_kN$, and since $f_k = 150\ N$ and $N = 470.4\ N$, we can find the coefficient of kinetic friction $\mu_k$.
$\mu_k=\frac{f_k}{N}$
$\mu_k=\frac{150}{470.4}\approx0.32$

Answer:

(a) $772.5\ J$
(b) $0.32$