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8. A 2 kg flower pot weighing 20 N falls from a window ledge.
How large must air resistance be in order that the pot fall
with an acceleration of 8 m/s2?

Respuesta :

onyebuchinnaji

The magnitude of the force of air resistance opposing the downward motion of the pot is 4 N.

The given parameters;

  • mass of the flower pot, m = 2 kg
  • weight of the flower pot, W = 20 N

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

  • This air resistance force decreases the magnitude of the downward force because it limits the downward motion of the pot as it falls from the window.
  • Let the air resistance = F

From Newton's second law of motion;

[tex]\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N[/tex]

Thus, the magnitude of the force of air resistance opposing the downward motion of the pot is 4 N.

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

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