Question

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?

Answer 1

Answer:

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.