Question

A sledge loaded with bricks has a total mass of 18.5 kg and is pulled at constant speed by a rope inclined at 19.5° above the horizontal. The sledge moves a distance of 19.3 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500.

(a) What is the tension in the rope?
N

(b) How much work is done by the rope on the sledge?
kJ

(c) What is the mechanical energy lost due to friction?
kJ

Answer 1

(a)

noraml force Fn = mg - T*sintheta

friction force fk = uk*Fn = uk*(m*g - T*sintheta)

along horizontal

acceleration ax = 0

Fnet = 0

Tx - fk = 0

T*costheta - uk*(m*g - T*sintheta) = 0

T*cos19.5 - 0.5*(18.5*9.81- T*sin19.5) = 0

tension T = 81.8 N

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(b)

work done by rope = T*costheta*L = 81.8*cos19.5*19.3 = 1488.2 J

Wrope = 1.50 kJ

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(c)

mechanical energy lost E = fk*L

E = uk*(m*g - T*sintheta)*L

E = 0.5*(18.5*9.81- 81.8*sin19.5)*19.3

E = 1.5 kJ