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| Saturday, September 8, 2012

I would like to mention that we do have a Facebook account and page. Please add the blog on Facebook (smartphysics39),  and like the page (smartphysics.tk) 


Furthermore, please follow the blog on Twitter (@SmartPhysicsTk).


That’s it for this short post. Thanks for reading!



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Forces and Newton's Laws of Motion Answers to Question 4

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Since the friend stops pulling on the box, and friction is now present, let’s make another diagram.



The box was moving before it hits the rough section. When the friend let go, the boy’s push and friction are suddenly equal, but in opposite directions. This is NOT an application of Newton’s third law of motion, since the forces are of different types.


So we go to Newton’s first law of motion. Since the forces are equal in magnitude and directed opposite to each other, they cancel out. The box then moves at a constant velocity. In other words, it keeps moving at a constant speed in one direction.


EXTRA: If the boy stop pushing as well, the box would decelerate (reduce speed) since the frictional force is directed opposite to the motion. This means that the acceleration is also directed in the opposite way, so the velocity should decrease in magnitude.



Forces and Newton's Laws of Motion Answers to Question 4

Forces and Newton's Laws of Motion Answers to Question 3

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The action-reaction pairs are:



  • The boy’s push (normal contact force) on the box, and the box’s push (also a normal contact force) on the boy.

  • The friends’s pull (through the string, called tension) on the box, and the box’s pull (also through the string, tension) on the boy.

  • The gravitational force on the box from the earth, and the gravitational force on the earth from the box.

  • The normal contact force on the box from the table, and the normal contact force on the table from the box.


Notice that all the forces in each action-reaction pair are of the same type. This is true for any action-reaction pair.



Forces and Newton's Laws of Motion Answers to Question 3

Forces and Newton's Laws of Motion Answers to Question 2B

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The forces are acting at perpendicular directions, as seen below.



If we use a vector diagram, we get the following:



Using the Pythagorean theorem,


F^{2}=(1.0 N)^{2} + (1.5 N)^{2}


F^{2}=3.25 N^{2}


F=\sqrt{3.25}N


F=1.80 N


Then the angle (which tells us the direction) can either be calculated or measured. If measure, the angle will be around 55 degrees to the left of the boy.



Forces and Newton's Laws of Motion Answers to Question 2B

Forces and Newton's Laws of Motion Answers to Question 2A

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The boy pushes from behind, and the friend in the opposite side, pulling.



Since the forces line up and are in the same direction, we add them up as:


\Sigma F = ma


1.0 N + 1.5 N = 2.0 kg \times a


a = \frac{2.5 N}{2.0 kg}


a = 1.25 ms^{-1}



Forces and Newton's Laws of Motion Answers to Question 2A

An Insider: Video and Slide Show Discussing the Basics of Vectors

| Tuesday, September 4, 2012

This post is intended for my students learning about vectors, without any problems. If you’re interested, you are more than welcome to download the files. 



Here’s the slide show, and here’s the video explanation. If you have any questions, feel free to post them on the comments section, and I’ll get to them as soon as I can.  :lol:


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An Insider: Video and Slide Show Discussing the Basics of Vectors

Test Post from School Science Site

| Sunday, September 2, 2012

Test Post from School Science Site http://www.schoolsciencesite.com