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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.

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Chenk13

Answer:

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:

[tex]a = - \frac{F}{m_1+m_2}[/tex]

[tex]N = -am_1[/tex]

case A:

|a| = 0.96 m/s²

|N| = 2.9 N

case B:

|a| = 0.644 m/s²

|N| = 3.86 N

onyebuchinnaji

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;

  • Case A:
  • mass of each block = 3.0 kg
  • Case B:
  • mass of block 1 = 6 kg
  • mass of block 2 = 3 kg
  • frictional force in both cases = 5.8 N

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

F = ma

5.8 = a(3 + 3)

5.8 = 6a

[tex]a = \frac{5.8}{6} \\\\a = 0.967 \ m/s^2[/tex]

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

[tex]a = \frac{5.8}{9} \\\\a = 0.64 \ m/s^2[/tex]

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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