1. Fp = Mg*sin24 = 0.407Mg.
Fn = Mg*Cos24 = 0.914Mg.
Fk = u*0.914Mg.
Fp-Fk = M*a.
0.407Mg-u*0.914Mg = M*a.
1.A block accelerates at 3.1 m/s2 down a plane
inclined at an angle 24.0◦.
Find μk between the block and the in-
clined plane. The acceleration of gravity is
9.81 m/s2 .
2.A 1200 kg car moves along a horizontal road
at speed v0 = 20.1 m/s. The road is wet,
so the static friction coefficient between the
tires and the road is only μs = 0.189 and
the kinetic friction coefficient is even lower,
μk = 0.1323.
The acceleration of gravity is 9.8 m/s2 .
What is the shortest possible stopping dis-
tance for the car under such conditions? Use
g = 9.8 m/s2 and neglect the reaction time of
the driver.
Answer in units of m.
Thank you for the help!
5 answers
Divide by Mg:
0.407-0.914u = a/g = 3.1/-9.81,
0.407-.914u = -0.316, uk = 0.791.
0.407-0.914u = a/g = 3.1/-9.81,
0.407-.914u = -0.316, uk = 0.791.
2. M*g = 1200*9.8 = 11,760 N. = Wt. of car. = Normal force(Fn).
Fp = Mg*sin 0 = 0.
Fk = uk*Fn = 0.1323*11,760 = 1556 N.
Fp-Fk = M*a.
0-1556 = 1200a, a = -1.30 m/s^2.
V^2 = Vo^2 + 2a*d.
0 = (20.1)^2 - 19.6d, d = ?.
Fp = Mg*sin 0 = 0.
Fk = uk*Fn = 0.1323*11,760 = 1556 N.
Fp-Fk = M*a.
0-1556 = 1200a, a = -1.30 m/s^2.
V^2 = Vo^2 + 2a*d.
0 = (20.1)^2 - 19.6d, d = ?.
Fp = Force parallel to the incline.
Fn = Normal force.
Fn = Normal force.
wth is this garbage
1 answer