A box with mass 1.10 kg is located on a horizontal tabletop with friction. The coefficient of kinetic friction is 0.500. The tabletop is square and measures 1.00 m on its side. The box starts at one corner and finishes at the diagonal edge. The path it follows is by first traveling one edge, turning and traveling to the final location. You push the box by exerting a force on it that makes an angle of 30.0o with the horizontal. How much work does the friction force do

W = 6.5 W

Explanation:

Work is defined by

         W = F . d

          W = f d cos tea

where the point represents the scaled product and the bold letters indicate vectors

they ask us the work of the friction force

we write the translational equilibrium equation

y Axis

                  N -W = 0

                  N = mg

x axis

                  F - fr = 0

                  F = fr

the formula for the friction force is

                  fr = μ N

we substitute

                  fr = μ m g

we substitute in the equation of work

           W = fr d cos θ

            W = μ m g d cos θ

let's calculate

            W = 0.500 1.10 9.8   Σ d_i  cos θ_i

            W = 5.39 d cos tea

we have two displacement

the first on one side of the box, suppose that side is on the y-axis, therefore the angle between the displacement and the friction force is 70º

and there is a second displacement in the x axis, in this case the angle between the friction force and the displacement is 30º

therefore the total workload is the sum of those work

            W = 5.39 (1 cos 70 + 1 cos 30)

             W = 5.39 (0.342 + 0.866)

             W = 6.5 W