Question

Mary applies a force of 77 N to push a box with an acceleration of 0.45 m/s2. When she increases the pushing force to 82 N, the box's acceleration changes to 0.74 m/s2. There is a constant friction force present between the floor and the box.
(a) What is the mass of the box?

Answer 1

Fap-Fk = m*a

Eq1: 77 - Fk = m*0.45
Eq2: 82 - Fk = m*0.74
Subtract Eq2 from Eq1:
Diff. = -5 = -29m
m = 0.172 kg.