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Three blocks are at rest on a smooth, horizontal table. You push on block A (m = 1.50 kg) to the right with a horizontal force of F push. As a result, block A pushes on block B (m = 2.50 kg), block B pushes on block C (m = 1.20 kg), and the system of blocks moves to the right with an acceleration of 1.12 m/s2. What is the magnitude of the force that block B exerts on block C?

Respuesta :

onyebuchinnaji

The magnitude of the force that block B exerts on block C is 4.48 N.

The given parameters;

  • mass of block A = 1.5 kg
  • mass of block B = 2.5 kg
  • mass of block C = 1.2 kg
  • acceleration of the three blocks = 1.12 m/c

The force block B exerted on block C is calculated from Newton's second law of motion;

[tex]F = ma\\\\\Sigma F = ma_{A} + ma_{B} + ma_{C}[/tex]

The force block B exerted on block C is also a combination of the force of block A.

The acceleration of block A and B only is calculated as;

[tex]a_{AB} = \frac{Ma_{A} + Ma_{B} }{M_A + M_B} \\\\a_{AB} = \frac{(1.5 \times 1.12) + (2.5\times 1.12)}{1.5 + 2.5} \\\\a_{AB} = 1.12 \ m/s^2[/tex]

The force exerted by both blocks on block C;

[tex]F_{\ B \ on \ C} = 1.12(1.5 + 2.5) \\\\F_{\ B \ on \ C} = 4.48 \ N[/tex]

Thus, the magnitude of the force that block B exerts on block C is 4.48 N.

Learn more here: https://brainly.com/question/19887955

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