What is the speed of a mass on a spring at a displacement of 0.2 m?

[CabalGroupie]

Homework Statement

A 2.0 kg mass on a spring is extended 0.3 m from equilibrium position and released. The spring constant is 65 N/m.

C.Find the speed of the mass when the displacement is 0.2 m

Homework Equations

Ek=(0.5)(m) (v^2)
Ep=(0.5)(k) (x^2)
Ek+Ep=E

The Attempt at a Solution

I've already ascertain from the previous parts of this question that the initial potential energy of the spring is 2.925 J, and the maximum speed reached by this mass is 1.71 m/s. According to my textbook the answer is 1.27 m/s but I keep getting...

(1/2)(m) (v^2)=(1/2)(k) (x^2)
v^2=(1/2)(65)(0.2^2)
v=1.14

 

[Panphobia]

The spring is initially stretched 0.3m from equilibrium, and then when it is released and is displaced 0.2m it is still stretched but has some velocity. So what is the energy before release, and what is the energy at 0.2m from equilibrium after release?

 

[Mentz114]

I solved this for v and got 1.27. You might have made a mistake.

(1/2)(m) (v^2)+(1/2)(k) (x^2)=(1/2)(m)(1.71^2)

[Edit: you should have used 0.3, not 0.2 here

(1/2)(m) (v^2)+(1/2)(k) (x^2)v^2=(1/2)(65)(0.3^2)

 

[haruspex]

(1/2)(m) (v^2)+(1/2)(k) (x^2)v^2=(1/2)(m)(1.71^2)

The second term is dimensionally wrong. I guess you meant (1/2)m v²+(1/2)k x²=(1/2)m1.71², though it's probably better to use (1/2)m v²+(1/2)k x₁²=(1/2)k x₀², minimising the growth of rounding errors.

 

[Mentz114]

The second term is dimensionally wrong. I guess you meant (1/2)m v²+(1/2)k x²=(1/2)m1.71², though it's probably better to use (1/2)m v²+(1/2)k x₁²=(1/2)k x₀², minimising the growth of rounding errors.

Sorry that v^2 is a typo. Tsk... I have corrected the post, thanks for pointing it out.