Solving for Maximum Speed: Object on Spring with Force Constant of 19.6N/m

[mizzy]

Homework Statement

A 0.40kg object connected to a light spring with a force constant of 19.6N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest, determine (a) the maximum speed of the object.

Homework Equations

F = -kx

The Attempt at a Solution

Does this involve kinetic energy?

 

[kjohnson]

Yes, you need to use the conservation of mechanical energy principle.

 

[mizzy]

so KE + KEs = 0

1/2mv^2 + 1/2kx^2 = 0

solve for v?

 

[kjohnson]

Close, but the spring has no kinetic energy only the attached mass. The spring can store what type of energy? Also set it up so its all the energy at state 1 equal to all the energy at state 2 (the point of interest).

 

[mizzy]

sorry not kinetic energy but potential energy.

 

[kjohnson]

Yes so you should arrive at:
PEs = KE
which would give the same result as if you used the work kinetic energy theorm:
Ws = KE2-KE1
where Ws is the work done by the spring on the block and is equal to the integral of the spring force with respect to x or:
Ws = integral of Fdx = (1/2)kx^2

 

[mizzy]

Yes so you should arrive at:
PEs = KE
which would give the same result as if you used the work kinetic energy theorm:
Ws = KE2-KE1
where Ws is the work done by the spring on the block and is equal to the integral of the spring force with respect to x or:
Ws = integral of Fdx = (1/2)kx^2

so 1/2mv^2 = 1/2kx^2
v = square root (kx^2/m)

 

[kjohnson]

correct.

 

[mizzy]

k. i got the answer, the max speed is 0.28m/s.

The next part of the question is, for what value of x does the speed equal one-half the maximum speed.

What i did was to solve for x with v = 1/2(max speed)

I got the answer wrong.

Can someone point out where i went wrong?

 

[kjohnson]

Again use the conservation of energy. Is your energy at state 1 (the compressed state) any different from part a? Then what forms of energy are present at state 2?

 

[mizzy]

at state 2, is it half the compressed state?

 

[kjohnson]

No, that distance x is the unknown you are asked to solve for. At state 1 the only energy is potential energy which you know. At state two you know the kinetic energy of the mass because you know v=vmax/2, but you don't know the potential energy of the spring=(1/2)kx^2...

 

[mizzy]

So in state 1 there is potential energy, no kinetic energy. State 2 there's kinetic energy, where v is half of the max speed and there's potential energy and that's when we solve for x.

 

[kjohnson]

correct.

 

[mizzy]

Thanks for you help! =)