Asked by Anonymous
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
Answers
Answered by
Henry
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
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
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