Question

1. -/2 points a box with a mass of 37.0 kg is pushed with a horizontal force of 150 n. the box moves at a constant speed across a level, rough floor a distance of 5.45 m. (a) what is the work done (in j) by the 150 n force? (b) what is the coefficient of kinetic friction between the box and the floor? resources read it

Explanation:

Step1: Calculate work - done by force

The work - done formula is $W = Fd\cos\theta$. Here, $F = 150\ N$, $d=5.45\ m$, and $\theta = 0^{\circ}$ (since the force and displacement are in the same direction, $\cos\theta=\cos0^{\circ}=1$). So, $W = 150\times5.45\times1$.
$W = 817.5\ J$

Step2: Analyze forces for coefficient of friction

Since the box moves at a constant speed, the net force in the horizontal direction is zero. So, the frictional force $f$ is equal to the applied force $F$, i.e., $f = 150\ N$. The normal force $N$ on the box on a level surface is equal to its weight $mg$, where $m = 37.0\ kg$ and $g = 9.8\ m/s^{2}$. So, $N=mg=37.0\times9.8 = 362.6\ N$. The formula for the frictional force is $f=\mu_{k}N$, and we can solve for the coefficient of kinetic friction $\mu_{k}$. Rearranging gives $\mu_{k}=\frac{f}{N}$. Substituting $f = 150\ N$ and $N = 362.6\ N$, we get $\mu_{k}=\frac{150}{362.6}\approx0.414$.

Answer:

(a) $817.5$
(b) $0.414$