A cart loaded with bricks has a total mass of 26.9 kg and is pulled at constant

Question

A cart loaded with bricks has a total mass of 26.9 kg and is pulled at constant speed by a rope. The rope is inclined at 25.5 degree above the horizontal and the cart moves 24.3 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.6. The acceleration of gravity is 9.8 m/s^2. What is the normal force exerted on the cart by the floor? Answer in units of N. How much work is done on the cart by the rope? Answer in units of kJ. What is the energy change W_f due to friction? Answer in units of kJ.

Explanation / Answer

Since there is no motion along the vertical direction ,then net force along vertical direction

Fnet,y = 0

F*sin(theta) + N -mg =0

F*sin(theta) = mg - N........(1)

and the cart is moving with constant speed ,accelaration along horizontal is ax = 0

F*cos(theta) = fk = mu_k*N .......(2)

(1)/(2)

tan(theta) = (N-mg)/(mu_k*N) = (1/mu_k) - (mg/(mu_k*N))

tan(25.5) = (1/0.6) - ((26.9*9.8)/(0.6*N))

Normal force is N = 369.32 N

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Work done by the rope is W = F*cos(theta)*S

F*cos(25.5) = mu_k*N = 0.6*369.32 = 221.6 N

then W = 221.6*24.3 = 5384.88 J

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Wf = W = 5384.88 J = 5.384 kJ

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