Two blocks are sliding to the right across a horizontal surface, as the drawing shows. in case a the mass of each block is 3.0 kg. in case b the mass of block 1 (the block behind) is 6.0 kg, and the mass of block 2 is 3.0 kg. no frictional force acts on block 1 in either case a or case b. however, a kinetic frictional force of 5.8 n does act on block 2 in both cases and opposes the motion. for both case a and case b determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.

The magnitude of the force with which the two blocks push against each other in cases A is 2.9 N and case B is 3.84 N.

The magnitude of the acceleration of the blocks in case A is 0.967 m/s². The magnitude of the acceleration of the blocks in case B is 0.64 m/s².

The given parameter;

The magnitude of the acceleration of the blocks in case A;

F = ma

5.8 = a(3 + 3)

5.8 = 6a

The magnitude of the force with which the two blocks push against each other in case A is calculated as;

F = 3 x 0.967

F = 2.9 N

The magnitude of the acceleration of the blocks in case B;

5.8 = a(3 + 6)

5.8 = 9a

The magnitude of the force with which the two blocks push against each other in case B is calculated as;

F = 6 x 0.64

F = 3.84 N

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Explanation:

According to a free body diagram the forces in the horizontal direction on body 1 would be:

F₁ = a₁*m₁ = -N

and on body 2:

F₂ = a₂*m₂ = N - F

N: normal force between the two blocks

F: frictional force on block 2

Since the two blocks are moving together, they need to have the same acceleration:

a₁ = a₂

This gives two equations with two unknown. Solving for a and N gives:

case A:

|a| = 0.96 m/s²

|N| = 2.9 N

case B:

|a| = 0.644 m/s²

|N| = 3.86 N

the answer is a. at the equator