A 80-kilogram rack star performs a workout preparing for a pole vault competition. The athlete starts from a rest on a horizontal surface and uniformly accelerates towards the bar at a top speed of 8.0 meters per second. The athlete sticks the pole into the ground and travels a vertical height of 5.7 meters and just clears the bar. Calculate the work done by friction by the athlete during the vault

athlete  W = 2560 J

roce      W_{fr} = - 2560 J

Explanation:

For this exercise we can use the relationship between work and kinetic energy

          W = ΔK

          W = K_f -K₀

as the bearing of rest its v₀ = 0 and the final velocity is v = 8 m / s

athlete's job is

           W = ½ m - 0

let's calculate

           W = ½ 80 8²

            W = 2560 J

For the work of the friction force let us use that since the athlete goes at constant speed, they relate it to zero, therefore the forces are in equilibrium

             ∑F = 0

             F- fr = 0

             F = fr

Therefore, the magnitude of the friction force is equal to the force applied by the athlete, as the friction force always has the opposite direction to the movement, its work is

             W_{fr} = - W_{atlete}

             W_{fr} = - 2560 J

answer: electrons

c

explanation: