Physics
posted by P on .
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.

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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