Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 8.9 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
Question
Answer:
Explanation:
The force of friction acting on the system
= .04 x 9.8 ( 10.2 + 7 )
= 6.74 N
Net force = 8.9 – 6.74
= 2.16 N
Acceleration in the system
= 2.16 / ( 10.2 + 7 )
= .12558 m / s ²
Contact force between boxes = FP
Considering force on box A
Net force = 8.9 – FP
Applying Newton’s law on box A
8.9 – FP = 10.2 x .12558
= 1.28
FP = 8.9 – 1.28
= 7.62 N