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Potential energy and a skateboard, only have until 11:30 
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#1
Feb2505, 08:48 PM

P: 17

A 54.5 kg. skateboarder starts out with a speed of 1.95 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does 265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.10 m/s
Calculate the change (PE = PEf  PE0) in the gravitational potential energy (b) How much has the vertical height of the skater changed? ok so potential energy is PE=MGH buti don't see a vertical height here! Please help!! 


#2
Feb2505, 09:08 PM

P: 281

right...
You can define the height to be zero anwhere you like. I would suggest letting the height that he starts at be zero. Set up your Conservation of Energy equations (ME_i = ME_f, where ME = KE +PE) using variables for any unknowns. 


#3
Feb2505, 09:14 PM

Sci Advisor
P: 412

= 103.62 + PE0 {Final Total Energy} = (1/2)*m*vf^2 + PEf = (1/2)*(54.5)*(6.10)^2 + PEf = = 1013.97 + PEf {Delta Total Energy} = {1013.97 + PEf}  {103.62 + PE0} = = 910.35 + {Delta PE} = = {Work Performed On Subject} = (80) + (265) = (185) {Delta PE} = (185)  (910.35) A) {Delta PE} = (1095.35 J) m*g*{Delta H} = (1095.35) {Delta H} = (1095.35)/{(54.5)*(9.8)} B) {Delta H} = (2.051 m) ~~ 


#4
Feb2505, 09:49 PM

P: 3

Potential energy and a skateboard, only have until 11:30
here's what i did. plug all the numbers into Wnc= (1/2mvf^21/2mv0^2)+(mghfmgh0) and you get Wnc= a number + a number *(h0hf)
then you go and find Wnc first since it already told you in the question= +80265 and you get 185. so 185=a number + a number *(h0hf) and you can now find out what (h0hf) equals to. which is close to 2. this is the answer to b (i got 2.15 cuz our variables are different on webassign) answer to a. since you know what h0hf is. change in PE= mghfmgh0 plug h0=hf+(your answer to B) and you get mghfmghfmg*(your answer to B) mghf cancels out. mg*(your answer to B) is the answer for A (negative number) hope that helps......LOL 


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