Find final speed using mechanical energy formula

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Homework Help Overview

The discussion revolves around a physics problem involving the conservation of mechanical energy, specifically calculating the final speed of a skateboarder at different points along a path. The original poster presents an equation based on energy conservation principles, noting an initial speed and height at two points.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the energy conservation equation and the calculations involved. There are questions regarding potential errors in mathematical steps and the handling of terms in the equations.

Discussion Status

Several participants have pointed out potential mathematical errors in the original poster's calculations. There is ongoing exploration of the equations used, with suggestions to simplify by canceling mass terms and to carefully re-evaluate the steps taken. The original poster indicates they are working through the feedback provided.

Contextual Notes

Participants note the assumption of negligible friction and the importance of correctly applying the conservation of mechanical energy principles. There is mention of a specific numerical error that may have affected the calculations.

Angela_vaal
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Homework Statement


A skateboarder at a skate park rides along the path shown in (Figure 1) .If the speed of the skateboarder at point A is v=1.3m/s, what is her speed at point B? Assume that friction is negligible.
fig_8-31.png


Homework Equations


ui+ki=uf+kf
u=mgh
k=1/2mv2

The Attempt at a Solution


m(9.8x2.7)+1/2m(1.3)2=m(9.8x1)+1/2mvf2
m(26.46+.845)=m(9.8+.5) mvf2
vf=√((26.46+.845)/(9.8+.5))
answer=1.6 m/s
The answer is wrong. Can somebody please tell me what I am doing wrong? Thanks!
 
Last edited by a moderator:
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Looks like you changed 0.845 to 8.45 from one step to the next. That could have caused a problem but I didn't verify that was it.
 
TomHart said:
Looks like you changed 0.845 to 8.45 from one step to the next. That could have caused a problem but I didn't verify that was it.

I wrote it down wrong here but I did the work using.845 and still got 1.6m/s. I don't know what I am doing wrong :(
 
I'll look at it again.
 
Yeah, you have a math problem in going from the first line to the second line. You should re-work that again carefully.
 
TomHart said:
Yeah, you have a math problem in going from the first line to the second line. You should re-work that again carefully.
I will do that right now.
 
I"m not that great at math. Can you give me a hint of what I did wrong?
 
@Angela_vaal you wrote:
m(9.8x2.7)+1/2m(1.3)2=m(9.8x1)+1/2mvf2
It's okay that you didn't, but at this point you could have (probably should have) canceled out all of the m's, because each term contains an 'm'.

Then you wrote:
m(26.46+.845)=m(9.8+.5) mvf2
I'm not exactly sure what you did to get to this equation. It looks like you're trying to factor mvf2 out of both terms on the right-hand side of the equation, but the first term does not contain a vf2. So that was not legal.

I think you ought to factor out all of the 'm's in the first equation, multiply out the numbers, and that will [hopefully] make it clearer to see how to proceed.
 
Angela_vaal said:

Homework Equations


ui+ki=uf+kf
u=mgh
k=1/2mv2

The Attempt at a Solution


m(9.8x2.7)+1/2m(1.3)2=m(9.8x1)+1/2mvf2
m(26.46+.845)=m(9.8+.5) mvf2
The RHS should be 9.8m+0.5 mvf2
Factor out m.
 
Last edited:
  • #10
Yeah I think it is just a math error as the general method of solving looks fine
 
  • #11
I figured it out thank you!
 

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