HELP Conservation of mechanical energy question

AI Thread Summary
The discussion centers on a problem involving the conservation of mechanical energy for a glider attached to a spring on a frictionless air track. The user initially calculates the speed of the glider incorrectly as 1.87 m/s instead of the correct answer of 1.47 m/s. Key insights reveal that the user mistakenly applied gravitational potential energy instead of spring potential energy in their calculations. The correct approach involves using the formula for spring potential energy (1/2 k x^2) to find the kinetic energy and subsequently the velocity. The conversation concludes with the user understanding the correct method to solve the problem.
abot
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I am having problems with this question. I keep getting an answer of 1.87m/s but the answer is 1.47m/s.

A glider of mass .150kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force of constant of 10 N/m both for extension and compression. The other end of the spring is fixed. The glider is moved to compress the spring by .180m and then released from rest. Calculate the speed of the glider (a) at the point where it has moved .180m from its starting point, so that the spring is momentarily exerting no force and (b) at the point where it has moved .250m from its starting point.
 
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abot said:
I am having problems with this question. I keep getting an answer of 1.87m/s but the answer is 1.47m/s.
Can't tell you where you went wrong if you don't show your work.
 
Wnc = K + U
0= .5(mv)^2 + mgh
v= sq.root{(2-(mgh)) / (m))
v= sq.root{2-(.15*9.8) / (.150)
v= sq.root(3.53)
v=1.87

I may be totally wrong but that's how i though it should be done.
 
You are you using gravitational PE--but you should be using spring PE (1/2 k x^2).

As the spring stretches, spring potential energy is transformed to kinetic energy. The air track is horizontal, so gravitational PE doesn't change.
 
yes i understand why you would use such equation but how would you get velecity from it
 
E_m_i = E_p + E_k
E_m_i = \frac{kx^2}{2} + 0

Plug and chug.

Do the same thing for E_m_f

Recall, this is an isolated system, therefore E_m_i = E_m_f

Solve for v
 
alright... got it...thanks
 
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