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 equilibrium condition

Since the object moves at a constant velocity, the pushing force \(F\) equals the kinetic - friction force \(F_f\). So \(F = F_f\).

Step2: Apply the formula for kinetic - friction force

The formula for kinetic - friction force is \(F_f=\mu_kN\), where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force.
Substituting \(F = F_f\) into \(F_f=\mu_kN\), we get \(N=\frac{F}{\mu_k}\).

Step3: Substitute the given values

Given \(F = 51.4\space N\) and \(\mu_k=0.472\). Then \(N=\frac{51.4}{0.472}\).

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

Answer:

\(109\)