Question

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?

Answer 1

Answer

given,

total mass = 18.3 Kg

inclined at = 19.8°

distance = 19.7 m

coefficient of friction = 0.5

a) Horizontal component of force forward,

        $F_x = F cos \theta$

Horizontal force backward = force of friction, f =μN.

Normal force, N = mg - F sinθ

Since the velocity is uniform, acceleration, a = 0

Net force, ΣF = 0

        F cos θ+ -μ(mg -F sinθ) = 0

        F (cosθ + μsin θ) = μmg

          $F = \dfrac{\mu m g}{cos \theta + \mu sin \theta}$

          $F = \dfrac{0.5\times 18.3\times 9.8}{cos 19.8^0 +0.5\times sin 19.8^0}$

                      =80.76 N

(b) Work done by the rope on the sledge,

        W = F.S cosθ = 80.76 x 19.7 x 0.9360

            = 1497 J

(c) Mechanical energy lost due to friction,

       E = W done against friction,

       W' = f  x S =F x S  = 80.76 x 19.7 x0.9360

            = 1497 J

we know velocity is uniform,.

       W done by the applied force = W done against friction.

           = 1440 J