Newton's Second Law and Elevator Motion

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Questions and Answers

In scenario 2, if the person has a mass of 70 kg and the elevator accelerates upwards at $2 m/s^2$, what is the normal force $N$ (in Newtons) acting on the person?

840 N

In scenario 3, describe what happens to the normal force if the downward acceleration, $a$, approaches $g$.

The normal force approaches zero.

Explain why the normal force ($N$) is zero in scenario 4.

In scenario 4, the elevator cable has been cut, and the elevator is in freefall. As the elevator accelerates downwards at $g$, the normal force becomes 0 because the person inside the elevator experiences weightlessness.

If the elevator is moving upwards at a constant velocity, what is the net force acting on the person inside the elevator? What is the normal force in this case, assuming the person's mass is $m$?

<p>The net force is zero, and the normal force is $mg$.</p> Signup and view all the answers

In scenarios 2 and 3, how does the perceived weight of the person change compared to their actual weight ($mg$)?

<p>In scenario 2 (elevator accelerating upwards), the person feels heavier, as the normal force is greater than $mg$. In scenario 3 (elevator accelerating downwards), the person feels lighter, as the normal force is less than $mg$.</p> Signup and view all the answers

Flashcards

Elevator at Rest/Constant Speed

When an elevator is stationary or moving at a constant velocity, the net force is zero, and the normal force equals the gravitational force.

Elevator Accelerating Upwards

When an elevator accelerates upwards, the normal force is greater than the gravitational force.

Elevator Accelerating Downwards

When an elevator accelerates downwards, the normal force is less than the gravitational force.

Elevator in Freefall

When an elevator's cable is cut, the elevator accelerates downwards at 'g'. The normal force becomes zero, resulting in weightlessness.

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Normal Force (N)

The upward force exerted by a surface (like an elevator floor) on an object, counteracting gravity.

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Study Notes

  • N represents a reading.

Acceleration = 0

  • The acceleration is zero.
  • N - mg = 0.
  • Therefore, N = mg.

Acceleration > 0 (Going Up)

  • The acceleration is greater than zero and the object moves upwards.
  • N - mg = ma.
  • Therefore, N = mg + ma.

Acceleration < 0 (Going Down)

  • The acceleration is less than zero and the object moves downwards.
  • mg - N = ma.
  • Therefore, mg - ma = N.

Free Fall

  • mg - N = mg.
  • Therefore, N = 0.

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