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1.0 kg block, a = 6 m/s². Applied force?

  1. A
    0 N
  2. B
    3 N
  3. C
    6 N
  4. D
    Cannot determine

Topic Flashcards

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Question

If a 1.0 kg block is accelerating at 6 m/s², what net force must be acting on it according to Newton's Second Law?

Answer

6 N (F_net = m*a = 1.0 kg * 6 m/s² = 6 N).

Question

In the equation F_net = m*a, what does "F_net" specifically represent? Why is this distinction important?

Answer

It represents the vector sum of all forces acting on the object. This is crucial because multiple forces (like friction, applied force, gravity) can act simultaneously; only their net value determines the acceleration.

Question

How would the calculated applied force change if the 1.0 kg block were on a frictionless horizontal surface versus being pushed vertically upward against gravity? Explain.

Answer

On a frictionless horizontal surface, the 6 N force is the applied force. To push it vertically upward at 6 m/s², the applied force must overcome gravity and provide the acceleration: F_app = m*(g + a) = 1.0 kg * (9.8 + 6) m/s² ≈ 15.8 N.

Question

What is the weight (gravitational force) of the 1.0 kg block on Earth? How does this compare to the 6 N net force accelerating it?

Answer

Weight = m*g ≈ 9.8 N. The accelerating net force (6 N) is less than the block's weight.

Question

If the same 6 N net force were applied to a 2.0 kg block, what would its acceleration be?

Answer

F_net / m = 6 N / 2.0 kg = 3.0 m/s².

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