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A sledge loaded with bricks has a total mass of 18.3 kg and is pulled at constant speed by a rope inclined at 19.8° above the horizontal. The sledge moves a distance of 19.7 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?
(b) How much work is done by the rope on the sledge?
(c) What is the mechanical energy lost due to friction?

Respuesta :

onyebuchinnaji

The tension in the rope pulling the sledge at constant speed is 80.78 N.  

The work done by the rope on the sledge is 1,496 J.

The mechanical energy lost due to friction is 1,496 J.

The given parameters;

  • mass of the brick, m = 18.3 kg
  • inclination of the rope, θ = 19.8⁰
  • distance traveled by the sledge , d = 19.7 m
  • the coefficient of friction, μ = 0.5

The normal force on the sledge is calculated as follows;

[tex]F_n = W - Tsin(\theta)\\\\F_n =(18.3\times 9.8) - Tsin(19.8)\\\\F_n = 179.34 -0.339T[/tex]

The frictional force on the sledge is calculated as;

[tex]F_k = \mu_k F_n\\\\F_k = 0.5(179.34 - 0.339T)\\\\F_k = 89.67 - 0.17T[/tex]

The net horizontal force on the sledge is calculated as follows;

[tex]Tcos(\theta) - F_k = 0\\\\Tcos(19.8) - (89.67 - 0.17T) = 0\\\\0.94T - 89.67 + 0.17T = 0\\\\1.11T = 89.67\\\\T = \frac{89.67}{1.11} \\\\T = 80.78 \ N[/tex]

The work done by the rope on the sledge is calculated as follows;

[tex]W = Tcos(\theta) \times d\\\\W = 80.78\times 0.94 \times 19.7 = 1,496 \ J[/tex]

The mechanical energy lost due to friction is calculated as follows;

[tex]W_f = F_k \times d\\\\W_f = (89.67 -0.17T)d\\\\W_f = (89.67 \ -\ 0.17\times 80.78) \times 19.7\\\\W_f = 1,496 \ J[/tex]

The energy lost due to friction and the work done by the rope are equal. This is the reason behind the constant speed of the sledge because the acceleration is zero.

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