Question

a box weighing 450 n is pulled along a level floor at constant speed by a horizontal rope. if the tension in the rope is 260 n, find the coefficient of kinetic friction.

Explanation:

Step1: Analyze the forces

Since the box is moving at a constant speed, the net force is zero. The horizontal force (tension \(T = 260N\)) equals the kinetic - friction force \(F_f\). The normal force \(N\) equals the weight of the box \(W=450N\) (because there is no vertical acceleration).

Step2: Use the formula for kinetic - friction

The formula for kinetic - friction is \(F_f=\mu_kN\). We know \(F_f = T=260N\) and \(N = W = 450N\). Rearranging the formula for \(\mu_k\), we get \(\mu_k=\frac{F_f}{N}\).

Step3: Calculate \(\mu_k\)

Substitute \(F_f = 260N\) and \(N = 450N\) into the formula \(\mu_k=\frac{260}{450}\approx0.58\)

Answer:

\(\mu_k\approx0.58\)