Question

alex is asked to move two boxes of books in contact with each other resting on a rough floor. he decides to move both of them at the same time by pushing on box A with a horizontal pushing force of Fp=8.9 N. here A has a mass of Ma= 11.0 kg and B has a mass of Mb= 7.0 kg. the contact force between the two boxes is Fc. the coefficient of kinetic friction between the boxes and floor is .02. (assume Fp acts in the +x direction)a) what is the magnitude of the acceleration of the two boxes? m/s^2b) what is the force exerted on Mb by Ma. In other words what is the magnitude of the contact force Fc? Nc) if alex were to push from the other side on the 7.0 kg box, what would the new magnitude of Fc be?

Answer 1

Answer:

Explanation:

Total frictional force on boxes

= total weight x coefficient of friction

= ( 11 + 7 ) x 9.8 x .02 = 3.53 N  

Net force on boxes

= 8.9 - 3.53

= 5.37 N

acceleration = 5.37 /( 11 + 7 )

= 0.3 m / s ²

b ) Let us consider movement of block A ( 11 kg  )

acceleration  a = .3 m/s²

friction force on block A

11 X 9.8 X .02

= 2.156 N

Net force on block A

8.9 - friction force - reaction force by block B

= 8.9 - 2.156 - F_ c

force = mass x acceleration

8.9 - 2.156 - F_ c = 11 x .3

F_ c  = 3.444 N

c )

If force is applied from the side of box B

We consider all forces on box B

frictional force on it

= 7 x 9.8 x .02

= 1.372 N

Net force on it

8.9 - 1.372 - F_c

force = mass x acceleration

= 8.9 - 1.372 - F_c = 7 x .3 ( Acceleration of box B will be the same that is 0.3 m/s² )

F_c = 5.428 N