A 52.9-kg skateboarder starts out with a speed of 2.04 m/s. He does 116 J of work on himself by pushing with his feet against the ground. In addition, friction does -264 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6 m/s. a. Calculate the change in the gravitational potential energy. b. How much has the vertical height of the skater changed, and is the skater above or below the starting point?

a. 1222.13 J b. 2.36 m. His new position is above his original position

Explanation:

From work-kinetic energy principle, workdone = change in kinetic energy

So work the skateboarder does on himself W₁ = -116 J

work done by friction W₂ = -264 J

gravitational potential energy change W₃ = mgΔy

The kinetic energy change ΔK = 1/2m(v² - u²) where m = mass of skater = 52.9 kg, u = initial speed of skaterboarder = 2.04 m/s and v = final speed of skaterboarder =  6 m/s. ΔK = 1/2m(v² - u²) = ΔK = 1/2 × 52.9(6² - 2.4²) = 842.13 J

So, W₁ + W₂ + W₃ = ΔK

W₃ = ΔK - W₁ - W₂ = 842.13 J - (-116 J) - (-264J) = 842.13 J + 116 J + 264J = 1222.13 J

b. Since W₃ = mgΔy = 1222.13 J

Δy = W₃/mg = 1222.13/(52.9 × 9.8) = 1222.13/518.42 = 2.36 m

Since Δy > 0, his new position is above his original position.

are their multiple choices ?

explanation: