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Potential energy and a skateboard, only have until 11:30

by kpangrace
Tags: 1130, energy, potential, skateboard
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kpangrace
#1
Feb25-05, 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!!
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MathStudent
#2
Feb25-05, 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.
xanthym
#3
Feb25-05, 09:14 PM
Sci Advisor
P: 412
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

(a) Calculate the change (PE = PEf - PE0) in the gravitational potential energy

(b) How much has the vertical height of the skater changed?
{Initial Total Energy} = (1/2)*m*v0^2 + PE0 = (1/2)*(54.5)*(1.95)^2 + PE0 =
= 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)


~~

zizikaboo
#4
Feb25-05, 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^2-1/2mv0^2)+(mghf-mgh0) and you get Wnc= a number + a number *(h0-hf)
then you go and find Wnc first since it already told you in the question= +80-265 and you get -185. so
-185=a number + a number *(h0-hf) and you can now find out what (h0-hf) 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 h0-hf is.
change in PE= mghf-mgh0
plug h0=hf+(your answer to B)
and you get
mghf-mghf-mg*(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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