Find final speed using mechanical energy formula

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SUMMARY

The discussion focuses on calculating the final speed of a skateboarder using the mechanical energy formula. The skateboarder starts with a speed of 1.3 m/s at point A and the calculations involve gravitational potential energy and kinetic energy. The initial attempt yielded an incorrect final speed of 1.6 m/s due to a mathematical error in the transition between equations. Participants highlighted the importance of canceling mass (m) from both sides of the equation to simplify calculations and avoid errors.

PREREQUISITES
  • Understanding of mechanical energy concepts, including potential and kinetic energy.
  • Familiarity with the equations of motion, specifically ui + ki = uf + kf.
  • Basic algebra skills for manipulating equations and solving for variables.
  • Knowledge of gravitational acceleration (9.8 m/s²) and its application in physics problems.
NEXT STEPS
  • Review the principles of conservation of mechanical energy in physics.
  • Practice solving problems involving gravitational potential energy and kinetic energy.
  • Learn how to simplify equations by canceling common factors in physics calculations.
  • Explore common pitfalls in algebraic manipulation within physics problems to avoid errors.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of common calculation errors in energy problems.

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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