Asked by marr
an electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350 N amd the box accelerates at 1.2 m/s^2 [forward] for 5 seconds. the cable breaks and the box slows down and stops. calculate the coefficient of kinetic friction?
Answers
Answered by
Henry
Wb = mg = 125kg * 9.8N/kg = 1225N. =
Weight of box.
Fb = (1225N,0deg.).
Fp = Fh = 1225sin(0) = 0 = Force parallel to plane = hor. force.
Fv = 1225cos(0) = 1225N = Force perpendicular to plane = The normal.
Fn = ma = 125 * 1.2 = 150N = Net force.
Fn = Fap - Fp - Ff = 150N,
350 - 0 - Ff = 150,
Ff = 350 - 150 = 200N,
Ff = u*Fv = 200,
1225*u = 200,
u = 0.163 = Coefficient of friction.
Fap = Force applied.
Weight of box.
Fb = (1225N,0deg.).
Fp = Fh = 1225sin(0) = 0 = Force parallel to plane = hor. force.
Fv = 1225cos(0) = 1225N = Force perpendicular to plane = The normal.
Fn = ma = 125 * 1.2 = 150N = Net force.
Fn = Fap - Fp - Ff = 150N,
350 - 0 - Ff = 150,
Ff = 350 - 150 = 200N,
Ff = u*Fv = 200,
1225*u = 200,
u = 0.163 = Coefficient of friction.
Fap = Force applied.
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