Asked by kdv
A rescue worker pulls an injured skier lying on a toboggan (with a combined mass of 127 kg) across flat snow at a constant speed. A 2.53 m rope is attached to the toboggan at ground level, and the rescuer holds the rope taut at shoulder level. If the rescuer's shoulders are 1.65 m above the ground, and the tension in the rope is 148 N, what is the coefficient of kinetic friction between the toboggan and the snow?
Answers
Answered by
Henry
Fs = m*g = 127kg * 9.8N/kg = 1245 N. =
Wt. of skier and toboggan.
sin A = Y/r = 1.65/2.53 = 0.6522
A = 40.7o
Fk = u*mg-u*T*sinA
Fk = 1245u - u*148*sin40.7
Fk = 1245u - 96.5u = 1148u
148*cos40.7 - 1148u = m*a = m*0 = 0
1148u = 112.2
u = 0.098
Wt. of skier and toboggan.
sin A = Y/r = 1.65/2.53 = 0.6522
A = 40.7o
Fk = u*mg-u*T*sinA
Fk = 1245u - u*148*sin40.7
Fk = 1245u - 96.5u = 1148u
148*cos40.7 - 1148u = m*a = m*0 = 0
1148u = 112.2
u = 0.098
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