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Newton's Law of Motion


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1838472234_ScreenShot2021-06-13at7_15_04PM.thumb.png.f331a0f2283fe04ee9357839579c8c34.pngIn each of the figure shown, coefficient of friction between surfaces in contact is 0.20 and bodies started from rest. Determine (a) acceleration of each body (b) tension in the cord supporting each body (c) final velocity of each body after 3 sec and (d) distance covered by each body after 3 sec.

 
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21 hours ago, Yey said:

 

1838472234_ScreenShot2021-06-13at7_15_04PM.thumb.png.f331a0f2283fe04ee9357839579c8c34.pngIn each of the figure shown, coefficient of friction between surfaces in contact is 0.20 and bodies started from rest. Determine (a) acceleration of each body (b) tension in the cord supporting each body (c) final velocity of each body after 3 sec and (d) distance covered by each body after 3 sec.

 

Any high school student would be able to do this problem in their head, I am not helping you with your homework! Why Don't you try learning about it.

 

Edited by VictorMedvil
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  • 2 months later...

Oh, dear! Perhaps when I was a highschool student I could do that in my head but not any more!  Now I have to use paper and pen! 

Representing the weight on the slope by a downward vector of magnitude 5g N, we can draw a line from the tip of that vector perpendicular to the slope and a line from that to the weight, creating a right triangle with hypotenuse of "length" 5g and angle at the weight 60 degrees.  The magnitude of the component of force parallel to the slope is 5g cos(60)= (5/2)g.  The magnitude of the component of force perpendicular to the slope is  5g sin(60)= (5sqrt(3)/2)g.  Since the coefficient of friction is 0.2= 1/5, the force pulling that weight down the slope is (5/2)g- (sqrt(3)/2)g= ((5- sqrt(3))/2)g.

 

For the weight on top, the downward force is 2g N so the friction force is (2/5)g N.   The net force on the two weights is ((5- sqrt(3))/2)g.  Since force is mass times accelertion, and the total mass is 7 kg, the acceleration is ((5- sqrt(3))/14)g.   

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