Question

A force of 35.0 N is required to start a 6.0-kg box movingacross a horizontal concrete floor. (a) What is the coefficientof static friction between the box and the floor?(b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction.

Answer 1

Step-by-step explanation:

Frictional force acting on the box = 35 N

If μ₁ be the coefficient of static friction and μ₂ be the kinetic friction

μ₁ mg = 35

μ₁ = 35 / 6 x 9.8

= .6

b )

net force acting on box to produce acceleration of .6 m /s²

= 6 x .6 = 3.6 N

net force acting on the box when it is accelerating

= 35 - μ₂ mg

35 - μ₂ x 6 x 9.8 = 3.6

35 - μ₂ x 58.8 = 3.6

μ₂ = .53