A 56.0 kg skateboarder starts out with a speed of 1.70 m/s.?

Question

A 56.0 kg skateboarder starts out with a speed of 1.70 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.05 m/s.

(a) Calculate the change (ΔPE = PEf - PE0) in the gravitational potential energy.

(b) How much has the vertical height of the skater changed?

Answer 1

Change in kinetic energy = 0.5*56(6.05^2 - 1.7^2)

Work done = 80 - 265

(a) ΔPE = - 0.5*56(6.05^2 - 1.7^2) + (80 - 265) = 758.95 J

(b) Δh = ΔPE/mg = 1.38 m