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December 21, 2014

December 21, 2014

Posted by **ju** on Thursday, November 11, 2010 at 3:10pm.

When you push a 1.75- book resting on a tabletop, it takes 2.23 to start the book sliding. Once it is sliding, however, it takes only 1.45 to keep the book moving with constant speed.

What is the coefficient of static friction between the book and the tabletop?

What is the coefficient of kinetic friction between the book and the tabletop?

- physics -
**MathMate**, Thursday, November 11, 2010 at 4:22pmWhat did you get, and how?

- physics -
**ju**, Thursday, November 11, 2010 at 5:38pmi tried it many different ways. but basically i used fk=ukN . forexample for static friction (uk) the acceleration= o

so unN=ma

us(1.45)=1.75kg(0)

us=1.45 and i got the wrong

- physics -
**ju**, Thursday, November 11, 2010 at 5:42pm...answer**

then for kinetic friction, i am clueless how to find the acceleration since there is no speed given in the equation. i know there is somehting i am missing or that i am looking at this question in the wrong way.

the varible i think are 1.75kg for the weight of the book

and 2.23N to start the books motion and 1.45N to get it at constant speed

just clearning that up from the question

- physics -
**MathMate**, Thursday, November 11, 2010 at 6:04pmThe question does not ask for acceleration. It only reduced the pushing force once it starts so that the book moves at constant speed.

You do not indicate units, which can be a major cause of the problem.

Force and can be expressed in newtons (N), and mass in kg.

Weight due to a mass of m (kg) is mg (N).

Fk=μk*N ...(1)

is correct, except that take care that N is expressed in newtons using

N=mg

so

μk = Fk/N = Fk/(mg)

Fs=μs*N

μs = Fs/N = Fs/(mg)

See if you find the correct answer with the above information.

- physics -
**mathhelp**, Thursday, November 11, 2010 at 6:35pmok got it. thank you.

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