# Homework ?

i have tried to figure out what formula to use on this and cant seem to get it plz help

A spring having elastic constant 22.8 N/m is stretched to a length 15.1 cm. Suppose an object of mass 130 g is attached to the end of the spring. (a) How much kinetic energy will the system have after the spring is released and moves through its equilibrium position? (b) What will the velocity of the object be at this position?
thanks ron

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truckerron1 said:
i have tried to figure out what formula to use on this and cant seem to get it plz help

A spring having elastic constant 22.8 N/m is stretched to a length 15.1 cm. Suppose an object of mass 130 g is attached to the end of the spring. (a) How much kinetic energy will the system have after the spring is released and moves through its equilibrium position? (b) What will the velocity of the object be at this position?
thanks ron
I'm presuming the spring is horizontal?

If so, then we know that the spring potential energy at an extension x beyond equilibrium is $$S=(1/2)kx^2$$. So initially, the spring has energy $$S=(1/2)(22.8N/m)(0.151m)^2$$ (watch the units!). Now, the Work-Energy Theorem says $$S_i+K_i=S_f+K_f$$. Note that the initial K is zero since the mass is just starting its motion. At the final point, when the spring is at its equilibrium length, what is x? From here you should be able to find v.

-Dan

PS If the spring is vertical, then we have to track the gravitational potential energy in the Work-Energy Theorem as well!

topsquark said:
I'm presuming the spring is horizontal?

If so, then we know that the spring potential energy at an extension x beyond equilibrium is $$S=(1/2)kx^2$$. So initially, the spring has energy $$S=(1/2)(22.8N/m)(0.151m)^2$$ (watch the units!). Now, the Work-Energy Theorem says $$S_i+K_i=S_f+K_f$$. Note that the initial K is zero since the mass is just starting its motion. At the final point, when the spring is at its equilibrium length, what is x? From here you should be able to find v.
I don't think so. The spring is stretched to a length $$15.1 cm$$. It is not the deformation of the spring, so you can have the equation $$S=(1/2)(22.8N/m)(0.151m)^2$$

phucnv87 said:
I don't think so. The spring is stretched to a length $$15.1 cm$$. It is not the deformation of the spring, so you can have the equation $$S=(1/2)(22.8N/m)(0.151m)^2$$
I don't understand what you are trying to say, phucnv87. If your comment is about:
PS If the spring is vertical, then we have to track the gravitational potential energy in the Work-Energy Theorem as well!
I was referring to the second part of the problem, not the first.

-Dan