Question

1. a box weighing 382 n is pulled horizontally until it slides uniformly over a level floor. if the applied force is 122 n, what is the coefficient of friction between the box and the floor?

Explanation:

Step1: Analyze the forces

When the box slides uniformly, the applied force \(F\) is equal to the frictional force \(f\). So \(f = F=122\space N\). The normal force \(N\) on the box is equal to its weight \(W\) since it's on a level floor, so \(N = W = 382\space N\).

Step2: Use the formula for frictional force

The formula for frictional force is \(f=\mu N\), where \(\mu\) is the coefficient of friction. Rearranging for \(\mu\), we get \(\mu=\frac{f}{N}\).

Step3: Calculate the coefficient of friction

Substitute \(f = 122\space N\) and \(N = 382\space N\) into the formula: \(\mu=\frac{122}{382}\approx0.32\)

Answer:

\(0.32\)