mg sin theta - F = ma
(9.8 sin 30) - 2 / (9.8 cos 30) = 0.34
you are skiing down a slope angled at 30 degrees to the horizontal. Your acceleration is 2m/s^2. what is the coefficient of kinetic friction between you and the hill?
2 answers
M*g = M*9.8 = 9.8M = Wt. of skier.
Fp = 9.8M*sin30 = 4.9M. = Force parallel with slope.
Fn = 9.8M*Cos30 = 8.49M = Normal force.
Fk = u*Fn = u*8.49M.
Fp-Fk = M*a.
4.9M-u*8.49M = M*2 = 2M,
u*8.49M = 2.9M, u = 0.342.
Fp = 9.8M*sin30 = 4.9M. = Force parallel with slope.
Fn = 9.8M*Cos30 = 8.49M = Normal force.
Fk = u*Fn = u*8.49M.
Fp-Fk = M*a.
4.9M-u*8.49M = M*2 = 2M,
u*8.49M = 2.9M, u = 0.342.