a. d = 0.5a*t^2 = 1.92 m.
0.5a*1.9^2 = 1.92 m
1.81a = 1.92
a = 1.06 m/s^2
b. M*g = 2.84 * 9.8 = 27.8 N. = Wt. of
block.
Fp = 27.8*sin30 = 13.9 N. = Force parallel to the incline.
Fn = 27.8*Cos30 = 24.1 N. = Normal force
Fp-Fk = m*a
13.9-Fk = 2.84*1.06 = 3.01
-Fk = 3.01-13.9 = -10.9
Fk = 10.9 N. = Force of kinetic energy.
uk = Fk/Fn = 10.9/24.1 = 0.452
c. Fk = 10.9 N. See part b.
d. V^2 = Vo^2 + 2a*d
Vo = 0
a = 1.06 m/s^2
d = 1.92 m.
Solve for V.
A 2.84 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 1.92 m in 1.90 s.
(a) Find the magnitude of the acceleration of the block.
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 1.92 m.
m/s
1 answer