Question

A 63.3-kg skateboarder starts out with a speed of 1.67 m/s. He does 81.1 J of work on himself by pushing with his feet against the ground. In addition, friction does -280 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6.37 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

given
m = 63.3 kg
vi = 1.67 m/s
w_skater = 81.1 J
w_friction = -280 J
vf = 6.37 m/s

a) we know, workdone by extenral force = change in mechanical energy

w_skater + w_friction = (KEf + PEf) - (KEo + PEo)

w_skater + w_friction = (KEf - KEo) + (PEf - PEo)

PEf - PEo = w_skater + w_friction - (KEf - KEo)

= 81.1 + (-280) - (1/2)*63.3*(6.37^2 - 1.67^2)

= -1395 J <<<<<<-----Answer

b) the vertical height of the skater changed, h = |PEf - PEo|/(m*g)

= 1395/(63.3*9.8)

= 2.25 m <<<<<<-----Answer