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.

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?

Henry · Answer

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.