Question

question 8 1 pts a box moves across a level tabletop at a constant velocity. it experiences an applied force of 116 newtons to the right. what must be the magnitude of the force of kinetic friction acting on the object? previous next not saved submit quiz

Explanation:

Step1: Analyze the motion

Since the box moves at a constant velocity, according to Newton's first law, the net force acting on it is zero.

Step2: Set up the force equation

Let \(F_{applied} = 116\space N\) (to the right) and \(F_f\) be the force of kinetic friction. The net force \(F_{net}=F_{applied}-F_f\). Since \(F_{net} = 0\) (constant velocity), we have \(0=F_{applied}-F_f\).

Step3: Solve for \(F_f\)

Rearranging the equation \(F_f=F_{applied}\)

Answer:

\(116\space N\)