When you push a 1.81 kg book resting on a tabletop, it takes 2.25 N to start the book sliding. Once it is sliding, however, it takes only 1.51 N 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?
I'm trying to approach this problem by:
f(k)=k*N.
f=kmg
k=f/mg
Wrong approach?
2007-11-13 08:03:01 · 2 answers · asked by revernance 3 in Science & Mathematics ➔ Physics
that is the correct approach.
k(static) = 2.25/(1.81*9.81)=.1267
k(dynamic) = 1.51/(1.81*9.81)=.0850
2007-11-13 08:25:20 · answer #1 · answered by Flyer 4 · 0⤊ 0⤋
f = u N is correct
u (static)= f/mg= 2.25/(1.81 x 9.81)
u (static)= 0.127
similarly
u (kinetic)= 1.51 / (1.81 x 9.81)
u (kinetic)=0.085
2007-11-13 16:28:34 · answer #2 · answered by Edward 7 · 1⤊ 0⤋