I tried mu= A/ mg sin30 But it didn\'t work? Is there something I\'m doing wrong
ID: 2062417 • Letter: I
Question
I tried mu= A/ mg sin30 But it didn't work? Is there something I'm doing wrong? Or is there an easier way to do it? Please and Thank you!A 3.08 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 2.13 m in 1.60 s.
(a) Find the magnitude of the acceleration of the block.
1.66 m/s2
(b) Find the coefficient of kinetic friction between the block and the incline.
(c) Find the magnitude of the frictional force acting on the block.
N
(d) Find the speed of the block after it has slid a distance 2.13 m.
2.66 m/s
Explanation / Answer
S = ut + (1/2)at^2, u=0 started from rest
2.13 = (1/2)a(1.6)^2
a = 1.664m/s2
Gravitational force = mgsin30 = 3.08*9.8*0.5 = 15.092N
Let frictional force = f
Net force = 15.092 - f = ma = 3.08*1.664 = 5.125
f = 9.966N
coeff of kinetic friction =
mgcos30 = f
= 9.966/(3.08*9.8*0.866) = 0.381
Change in potential energy = workdone by friction + Kinetic energy
mgh = f*2.13 + (1/2)mv^2
3.08*9.8*2.13sin30 = 9.966*2.13 + (1/2)mv^2
(1/2)*3.08*v^2 = 10.916
v = 2.66m/s