A 61.4-kg skateboarder starts out with a speed of 2.31 m/s. He does 118 J of wor
ID: 1352519 • Letter: A
Question
A 61.4-kg skateboarder starts out with a speed of 2.31 m/s. He does 118 J of work on himself by pushing with his feet against the ground. In addition, friction does -253 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6.15 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.
A 61.4-kg skateboarder starts out with a speed of 2.31 m/s. He does 118 J of work on himself by pushing with his feet against the ground. In addition, friction does -253 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 6.15 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.Explanation / Answer
A)
net work done = chane in mechanical energy = chnage in KE + change in PE
change in PE = net work done - change in KE
change in PE = (118-253)-(0.5*61.4*(6.15^2 - 2.31^2))
change in PE = -135 - 32.48 =-167.48 J
B) m*g*h = 167.48
h = 167.48/(61.4*9.81) = 0.27 m
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