Answer to Question #95820 in Mechanics | Relativity for Logan Wadsworth

Question

A ship launched from a dry-dock slides into the water at a constant velocity. Suppose the force of gravity that pulls the ship downward along the dry-dock is 4.26 X 10’7 N. If the coefficient of kinetic friction between the ships hull and the dry dock is 0.25, what is the magnitude of the normal force hat the dry dock exerts on the ship’s hull?

Answer 1

$$W=\mu N$$

$$N=\frac{W}{\mu}=\frac{4.26\cdot 10^{7}}{0.25}=1.70\cdot 10^{8}N$$