Fc = m*g = 22.3kg * 9.8N/kg = 218.54 N = Wt. of cart.
Fk = u*Fc = 0.442 * 218.54 = 96.6 N. =
Force of kinetic friction.
Fap*cos27.3-Fk = m*a
Fap*cos27.3-96.6 = m*0 = 0
Fap*cos27.3 = 96.6 N.
W=Fap*cos27.3 * d = 96.6 * 9.3=898.3 J.
Fap = Force applied = Tension in rope.
A cart loaded with bricks has a total mass of 22.3 kg and is pulled at constant speed by a rope. The rope is inclined at 27.3◦degrees above the horizontal and the cart moves 9.3 m on a horizontal surface. The coefficient of kinetic friction between ground and cart is 0.442. The acceleration of gravity is 9.8 m/s^2. How much work is done on the cart by the rope?
Answer in units of J.
2 answers
Correction:
Fk = u*Fc-u*Fap*sin27.3
Fk = 0.442*218.54 - 0.442*Fap*sin27.3
Fk = 96.6 - 0.203Fap
Fap*cos27.3-(96.6-0.203Fap) = m*a
0.889Fap - 96.6 + 0.203Fap = m*0 1.092Fap = 96.6
Fap = 88.5 N. = Force applied.
W = Fap*cos27.3 * d
W = 88.5*cos27.3 * 9.3 = 731.3 J.
Fk = u*Fc-u*Fap*sin27.3
Fk = 0.442*218.54 - 0.442*Fap*sin27.3
Fk = 96.6 - 0.203Fap
Fap*cos27.3-(96.6-0.203Fap) = m*a
0.889Fap - 96.6 + 0.203Fap = m*0 1.092Fap = 96.6
Fap = 88.5 N. = Force applied.
W = Fap*cos27.3 * d
W = 88.5*cos27.3 * 9.3 = 731.3 J.