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

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?
1
Expert's answer
2020-05-18T10:07:58-0400

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"


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