A boy pulls a bag of baseball bats across a ball field toward the parking lot. The bag of bats has a mass of 6.8 kg, and the boy exerts a horizontal force of 20 N on the bag. As a result, the bag accelerates from rest to a speed of 1.12 m/s in a distance of 5.4 m.

Question

Part A
What is the coefficient of kinetic friction between the bag and the ground? 3sf

Answer 1

V^2 = Vo^2 + 2a*d.
1.12^2 = 0 + 2a*5.4, a = 0.116 m/s^2.

M*g = 6.8*9.8 = 66.64 N. = Wt. of bats. = Normal force(Fn).

Fk = u*Fn = 66.64u.

Fe-Fk = M*a.
20-66.64u = 6.8*0.116 = 0.789,
66.64u = 20-0.789 = 19.2,
u = 0.288.