Question

fill in the blank question
pushing with a force of 51.4 n allows an object on a surface with a coefficient of kinetic friction of 0.472 to move at a constant velocity. the normal force is equal to
n. round to three significant figures.

Explanation:

Step1: Use the kinetic friction formula

When the object moves at a constant velocity, the applied force \(F\) is equal to the kinetic - friction force \(F_f\). The formula for kinetic - friction force is \(F_f=\mu_kF_N\), where \(\mu_k\) is the coefficient of kinetic friction and \(F_N\) is the normal force. Since \(F = F_f\), we have \(F=\mu_kF_N\).

Step2: Solve for the normal force

We can re - arrange the formula \(F=\mu_kF_N\) to solve for \(F_N\). The formula becomes \(F_N=\frac{F}{\mu_k}\).
Given that \(F = 51.4\space N\) and \(\mu_k=0.472\), we substitute these values into the formula: \(F_N=\frac{51.4}{0.472}\).

$$F_N=\frac{51.4}{0.472}\approx109\space N$$

Answer:

\(109\)