Question #116371

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?

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?

Expert's answer

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"

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment