Posted by P on Wednesday, December 5, 2012 at 10:44am.
A 64.4kg skateboarder starts out with a speed of 2.11 m/s. He does 81.6 J of work on himself by pushing with his feet against the ground. In addition, friction does 289 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 8.56 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.

Physics  Che. It's yur boi!, Monday, November 30, 2015 at 6:27pm
ΔKE=1/2mvf^2 1/2mv0^2
a)Plug numbers in
1. ΔKE=1/2(64.4)(8.56)^2 1/2(64.4)(2.11)^2= 2216.05 (because displacement, I think)
2. 2216.05J+(289J)80J
Answer =2425.05J
b.)
1. W=ΔKE, W=ΔPE(PE=mgh), ΔPE=ΔKE
(64.4kg)(9.8m/s^2)h=2425.05J
631.12h=2425.05J
Answer h=3.842m
Hope this Help
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