Newton's 2nd Law of Motion Lab Review

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Questions and Answers

What is the acceleration of a 0.20 kg ball when a force of 3.6 N is applied to accelerate it upward?

  • 9 m/s$^2$
  • 12 m/s$^2$
  • 15 m/s$^2$ (correct)
  • 18 m/s$^2$

If a 0.20 kg ball is accelerated downward, what force is required to achieve an acceleration of 12 m/s$^2$?

  • 2.4 N
  • 3.6 N
  • 5.7 N (correct)
  • 4.8 N

When a 0.20 kg ball is in free fall, what is the net force acting on it?

  • 9.8 N downward
  • 0.20 N downward
  • 0 N (correct)
  • 9.8 N upward

What is the weight of a person experiencing an acceleration of 5 m/s$^2$ upward if the apparent weight is 700 N?

<p>500 N (D)</p> Signup and view all the answers

If the air resistance on a falling object increases while its weight remains constant, what happens to its acceleration?

<p>Decreases (A)</p> Signup and view all the answers

According to Newton's second law, what is the relationship between force and acceleration?

<p>Force is directly proportional to acceleration (A)</p> Signup and view all the answers

What is the relationship between mass and acceleration according to Newton's second law?

<p>Mass is inversely proportional to acceleration (D)</p> Signup and view all the answers

If an object is at rest and there is no net force acting on it, what will happen to the object according to Newton's laws?

<p>The object will remain at rest (A)</p> Signup and view all the answers

If an object is moving with a constant velocity and there is no net force acting on it, what will happen to the object according to Newton's laws?

<p>The object will continue moving at a constant velocity (A)</p> Signup and view all the answers

Which of the following is a correct free-body diagram for a car speeding up while moving to the right?

<p>A car speeding up while moving to the right (B)</p> Signup and view all the answers

If you are given data for a lab experiment, which of the following should you graph to create a linear fit according to the text?

<p>Force vs. Acceleration (C)</p> Signup and view all the answers

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