Posted by Stephanie on Saturday, October 15, 2011 at 12:55pm.
When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?

physics  Henry, Sunday, October 16, 2011 at 6:16pm
Wb = mg = 1.7kg * 9.8N/kg = 16.66N.
Fb = (16.66N,0deg) = Force of book..
Fp = 16.66sin(0) = 0 = Force parallel to table.
Fv = 16.66cos(0) = 16.66N. = Force perpendicular to table.
1. Fap  Ff = 0, Fap = Force applied.
2.60  Ff = 0, Ff = Force of friction.
Ff = 2.60,
u*Fv = 2.60,
16.66u = 2.60,
u = 0.156. = Static coefficient of friction.
2. 1.5  Ff = 0,
Ff = 1.5,
u*Fv = 1.50,
16.66u = 1.5,
u = 0.09 = Kinetic coefficient of friction.
Answer This Question
Related Questions
 physics  When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to ...
 physics  When you push a 1.55 kg book resting on a tabletop it takes 2.10 N to ...
 physics  When you push a 1.55 kg block resting on a tabletop it takes 1.85 N to...
 physics  When you push a 2.05 kg book resting on a tabletop it takes 2.45 N to ...
 physics  When you push a 1.95 kg book resting on a tabletop it takes 2.40 N to ...
 Physics  When you push a 1.82kg book resting on a tabletop, it takes 2.30 N to...
 phy  When you push a 1.82{\rm kg} book resting on a tabletop, it takes 2.21 N ...
 physics  I keep geting this question wrong and cant figure out what i am doing ...
 Physics  When you push a 1.80kg book resting on a tabletop, it takes 2.05 N to...
 Science  You push a book on a table so that its velocity is constant. If the ...
More Related Questions