Question

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

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




Related Questions