Answer to Question #116371 in Mechanics | Relativity for Michael Mateus

Question

A force of 48.0 N is required to start a 6.0 Kg box moving across a horizontal
concrete floor.
a) What is the coefficient of static friction between box and floor?
b) If the 48.0 N force continues, the box accelerates at 0.70m/s2
, what is the
coefficient of kinetic friction?

Answer 1

Based on Newton's second law

a) $ma=F-kmg$ as $a=0$ hence $k=\frac{F}{mg}=\frac{48}{8\sdot9.8}\approx0.6$$1$

b) as $a=0.7$ hence $k=\frac{F-ma}{mg}=\frac{48-8\sdot0.7}{8\sdot9.8}\approx0.5$$4$