Question

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0° below the horizontal. The coefficient of kinetic friction between the package and the chute’s surface is 0.40. Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package

Answer 1

Answer

given,

mass of the package = 12 kg

slides down distance = 2 m

angle of inclination = 53.0°

coefficient of kinetic friction = 0.4

a) work done on the package by friction is

              W_f = -μk R d

                   = -μk (mg cos 53°)(2.0)

                   =-(0.4)(8.0 )(9.8)(cos 53°)(2.0)

                   = -37.75 J

b)

work done on the package by gravity is

             W_g = m (g sin 53°) d

                   = (8.0 )(9.8 )(sin 53°)(2.0 )

                   =125.23 J

c)

the work done on the package by the normal force is

             W_n = 0

d)

the net work done on the package is

           W = -37.75 + 125.23 + 0

           W = 87.84 J