Question

54.0-kg skateboarder starts out with a speed of 2.38 m/s. He does 105 J of work on himself by pushing with his feet against the ground. In addition, friction does -230 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 8.53 m/s. (a) Calculate the change (PEf - PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.

Answer 1

Then Initial k.e.K =(1/2)*m *vi^2

=0.5*54.0*2.38^2

= 153 J

Final k.e.Kf =(1/2)*m*vf^2

=0.5*54.0*8.53^2

=1964.54 J

Change in k.e.AK=Kf - Ki

=1964.54-153

=1811.54 J

Work done on skater W= 105 J-230J=-125 J

Hence change in gravitational potentialenergy AU = K - W

m*g*h=1811.54-(-125)

54.0*9.8*h=1936.54

change in height h = 1936.54 / 529.2=3.659m

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