Asked by Johnna
A 3.00 kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.75 s. Find the coefficient of kinetic friction between the block and the incline
Answers
Answered by
Henry
Wb = mg = 3kg * 9.8N/kg = 29.4N.
Fb = (29.4N,33deg).
Fp = 29.4sin33 = 16.o1N. = Force parallel to incline.
Fv = 29.4cos33 = 24.66N. = Force perpendicular to incline.
d = Vo*t + 0.5at^2 = 2.0m,
0*1.75 + 4.9a*(1.75)^2 = 2,
15a = 2,
a = 0.133m/s^2.
Fn = Fp - Ff = ma,
16.01 - Ff = 3 * 0.133,
16.01- Ff = 0.40,
Ff = 16.01 - 0.40 = 15.61N.
u = Ff / Fv = 15.61 / 24.66 = 0.633 =
coefficient of kinetic friction.
Fb = (29.4N,33deg).
Fp = 29.4sin33 = 16.o1N. = Force parallel to incline.
Fv = 29.4cos33 = 24.66N. = Force perpendicular to incline.
d = Vo*t + 0.5at^2 = 2.0m,
0*1.75 + 4.9a*(1.75)^2 = 2,
15a = 2,
a = 0.133m/s^2.
Fn = Fp - Ff = ma,
16.01 - Ff = 3 * 0.133,
16.01- Ff = 0.40,
Ff = 16.01 - 0.40 = 15.61N.
u = Ff / Fv = 15.61 / 24.66 = 0.633 =
coefficient of kinetic friction.
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