A 2.5 -kg block slides down a 25 degree inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity reaches .65 m/s the lenght of the incline is 1.6m.
what is the coefficent of friction between the plane and the book?
2007-11-01 14:56:38 · 1 answers · asked by james 2 in Science & Mathematics ➔ Physics
block not book.
2007-11-01 15:08:31 · update #1
what is s?
2007-11-01 15:40:39 · update #2
How did the book enter the picture?
Since acceleration is constant
Fd - f = F
Fd= mg sin(A)
f= umg cos(A)
F=ma
I guess that should do it
and since initial velocity was 0
a=V/t
S=0.5 a t^2
S=0.5 (V/t) t^2
S= 0.5 V t
t= 2S/V
then a=V^2/2S lets go back to our force equation
Fd - f = F
mgsin(A) - umg cos(A) = m (V^2 / 2S)
gsin(A) - ug cos(A) = (V^2 / 2S)
u= [gsin(A) - (V^2 / 2S)]/ g cos(A)
That is it!
Have fun!
2007-11-01 15:05:56 · answer #1 · answered by Edward 7 · 0⤊ 0⤋