The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1x=x1. Find _f, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.) Express the frictional force in terms of mmm, v0v0v_0, and x1x1x_1.

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk = –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer