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Chapter 12 Review a d 13 kg b c A box is accelerated to the right across the smooth surface of a table with an applied force of 40 Newtons. The table surface.

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Presentation on theme: "Chapter 12 Review a d 13 kg b c A box is accelerated to the right across the smooth surface of a table with an applied force of 40 Newtons. The table surface."— Presentation transcript:

1 Chapter 12 Review a d 13 kg b c A box is accelerated to the right across the smooth surface of a table with an applied force of 40 Newtons. The table surface provides a friction coefficient of .2 Label the forces and their values at letters A-D. Determine the net force C.O.F. x Normal F = Friction 1 kg of mass = 10 Newtons

2 Identify the forces at each colored arrow: 5 forces total.

3 Question 1: Which way is the box moving?
Question 2: Why is the Normal force arrow shorter than the gravity arrow?

4 Chapter 12 Review Which of the following is a non-contact force?
A) Friction B) Applied C) Gravity D) Normal

5 1 kilogram = 10 newtons How much force would be needed to lift the box off of the table? 10 kg

6 What force will cause this diver to be propelled off of the diving board?
a) Tension b) Applied c) Spring d) Normal

7 Chapter 12 Review Graph 1 Q Identify each type of graph.
(what are they showing) Q. Describe the motion of the object in each graph and explain any changes in that objects motion. d (m) T (sec) Graph 2 V (m/s) T (sec)

8 Chapter 12 Review Segment Distance Time Ave. Speed A - B B - C C - D
A - D Complete the table above with the correct values.

9 Chapter 12 Review Calculate the the acceleration between segments B - C How far did the object travel between segments A – B? How many times did this object reach a constant speed?

10 Chapter 12 Review a Kevin applies a force of 150 Newtons
to the wooden crate and it does not move. What is the exact name and magnitude in Newtons of the force at letter B? c b The second time, Kevin applies a force of 200 Newtons and the box slowly slides across the floor. Is there still a force at letter B? Explain If there is a force at letter b, how could Kevin reduce that force? b

11 Answers Slide 1: Box on table Forces:
Forces:  Gravity letter C = N (13 kg x 9.8) F = ma Normal Force letter A = N (same as above) Applied Force B = 40 N Friction (kinetic) letter D = N (.2 x Normal F) Net Force = N Right Acceleration = 1.11 m/s/s Slide 2 Motion & Acceleration graphs Graph 1 is a speed graph. The motion shown is an object that starts out In motion, stops for a short time and then resumes its motion. Graph 2 is an acceleration graph. The object accelerates, then reaches a constant speed, this is followed by negative acceleration, and then the final segment shows the objet accelerating positively again.

12 Answers Slide 4 Acceleration 1. – 3.3 m/s/s meters 3. 3 times

13 Answers Slide 5 Friction forces Diagram 1
Diagram 1 Static friction (150 N) is = to Kevin’s applied force of 150 Newtons therefore the horizontal (sideways) forces are equal, thus no acceleration. Diagram 2 There is still a force present at letter b. That force is now kinetic or sliding friction. Since Kevin is applying greater force, he overcomes static friction, but still has to deal with the kinetic friction of the crate against the floor. He could reduce this by putting the crate on wheels. Rolling friction is many times less than sliding friction.


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