Ws = M*g = 9.8M.
Fp = 9.8M*sin30 = 4.9M.
Fn = 9.8M*Cos30 = 8.49M.
Fk = u*Fn = u*8.49M.
Fp-Fk = M*a.
4.9M-u8.49M = M*(-1.4)
Divide by M:
4.9-8.49u = -1.4
-8.49u = -1.4-4.9 = -6.3
u = 0.74.
An Olympic turkey moving at 20.0 m/s down a 30.0° slope encounters a region of
wet snow and slides 145 m before coming to a halt. What is the coefficient of friction
between the turkey and the snow? The accelerations down the hill is 1.4 m/s^2.
2 answers
Notes:
1. Fp = Force parallel with the hill.
2. Fn = Normal force.
3. Fk = Force of kinetic friction.
4. Ws = Wt. of skier.
1. Fp = Force parallel with the hill.
2. Fn = Normal force.
3. Fk = Force of kinetic friction.
4. Ws = Wt. of skier.
1 answer