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**

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