Glider on spring: finding speed from elastic potential energy

[abot]

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.

 

[Doc Al]

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.

 

[abot]

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.

 

[Doc Al]

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.

 

[abot]

yes i understand why you would use such equation but how would you get velecity from it

 

[Zlex]

$$E_m_i = E_p + E_k$$
$$E_m_i = \frac{kx^2}{2} + 0$$

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

 

[abot]

alright... got it...thanks