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Unformatted text preview: • The coefficient of static friction between a 12.5 kg block and table is 0.34 and the coefficient of kinetic friction is 0.22. The block is at rest on the table when a horizontal force of 30 N is applied to it. A). What is the force of friction at this point in time? B). The block is now set in motion (by temporarily applying a larger force), and then the same 30 N force is applied. What is the acceleration in this second case? For both questions give a brief description of how you arrived at your answer. (A: 8 correct 19 incorrect; B: 9 correct 18 incorrect; 28 none) i • a)10 N. This is calculated by multiplying the number of Newtons applied to the stationary block by the coefficient of friction. • A) F= mg*0.34=41.65 N (as block is not moving we are using coefficient of static friction) u • b) 2.1 m/s^2. This is calculated by multiplying 12.5 by 9.8 and the .22 coefficient of kinetic friction. • f s,max = 0.34*N = 0.34*12.5kg*9.8m/s 2 = 41.6N, since this is greater than the applied force f s =30N opposite the applied force (since a=0 the forces must cancel !!). • “The force of friction always opposes motion.” Please comment briefly on the validity and/or universality of this statement. (Correct: 5 Incorrect: 24 No answ: 24) ( • From Newton's 2nd law, a force must act oppposite of any given force, so when an object is pushed a force must push back against it in the direction opposite its velocity, this force is frictional force....
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This note was uploaded on 01/21/2012 for the course ECE 476 taught by Professor Overbye,t during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Overbye,T

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