h = 3.1*sin30 = 1.55 m.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.55 = 30.4
V = 5.51 m/s.
b. V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d = (0-(5.51^2))/10 =
-3.04 m/s^2.
M*g = 7.5 * 9.8 = 73.5 N. = Wt. of suitcase.
Fp = 73.5*Sin30 = 36.75 N. = Force
parallel to the incline.
Fn = 73.5*Cos30 = 63.7 N. = Force
perpendicular to the incline.
Fp-Fk = M*a
36.75-Fk = 7.5*-3.04 = -22.8
-Fk = -22.8-36.75 = -59.55
Fk = 59.55 N. = Force of kinetic friction.
u = Fk/Fn = 59.55/63.7 = 0.93
starting from rest a 7.5 kg suitcase slides down a 30 degree frictionless incline a distance of 3.1 m upon reaching the bottom it slides an additional 5 m before coming to a stop. a) determine speed of suitcase at bottom of ramp b) determine coefficient of kinetic friction between suitcase and floor c) determine change in mechanical energy of suitcase due to friction
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