Finding Maximum Speed of Block on Vertical Spring

[Thefox14]

Homework Statement

[PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-08-GPE-ME/mass_vertical_spring/7.gif

A spring with spring constant k = 35 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 1 kg is hung gently on the end of the spring.

a) How far does the spring stretch?

Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 3 m/s.
b) What is the maximum speed of the block?

Homework Equations

W_{Total} = \Delta K
W_{NC} = \Delta ME

The Attempt at a Solution

Part a was easy, got the answer to be ~.28m

Part b is where I'm having problems. So far this is what I've done:
I decided to use U₀ + K₀ = U_f + K_f and set U = 0 at the spot where the spring is at equilibrium with the mass on it. So I got this equation:
\frac{1}{2}kx^{2} - mgx + \frac{1}{2}mv_{0}^{2}= \frac{1}{2}mv_{f}^{2}

Where x is .28 (the additional stretch). I'm not getting the correct answer which is around 3.4m/s

Any ideas what's wrong?

 

[rl.bhat]

The mass will have maximum speed when it crosses the unstretched position of the spring.
So in the energy formula the displacement of the mass is 3x.

 

[Thefox14]

Sadly even If i use x = 3*.28m i don't get the answer of 3.4 m/s

 

[Quinzio]

You have to know that the block will swing simmetrically around it's rest point.

The spring will always have an "offset" length so that it will cancel the gravity force mg.
Then you can use the conservation laws.

 

[Thefox14]

You have to know that the block will swing simmetrically around it's rest point.

The spring will always have an "offset" length so that it will cancel the gravity force mg.
Then you can use the conservation laws.

I understand that it will oscillate between its rest point, but I'm just looking for the place where the kinetic energy is the greatest which would be at the point where the spring is relaxed with the block on it right? Not sure how i'd apply what you're saying

 

[Quinzio]

I understand that it will oscillate between its rest point, but I'm just looking for the place where the kinetic energy is the greatest which would be at the point where the spring is relaxed with the block on it right?

Ok

Not sure how i'd apply what you're saying

Use the last formula of the 1st post, but ignore mgx. You should understand why it has to be ignored.

 

[Thefox14]

Wow, yeah that gave me the right answer. I don't understand why you can ignore gravity though. Could you explain that a bit?

 

[Quinzio]

If you want to keep mgx in the equation, you should calculate the spring energy starting from its unstretched point (without the block).
In the end the cancel out each other, when block+spring are at rest. It's more intuitive to ignore it.

 

[Thefox14]

So just out of curiosity what would the equation be if I decided to keep mgx in?

Wouldn't it be the same equation I had above except x would be 2*.28m as now the stretch from U = 0 is .56m

 

[Quinzio]

So just out of curiosity what would the equation be if I decided to keep mgx in?

Wouldn't it be the same equation I had above except x would be 2*.28m as now the stretch from U = 0 is .56m

Sorry, I didn't see your post.

Start from your equation that I changed a bit, you can do on your own.

\frac{1}{2}kx^{2} + mgx + \frac{1}{2}mv_{f}^{2} + \frac{1}{2}mv_{0}^{2}= 0

This equation tell you that you defined the energy at the top point as zero, and must be zero in each point.

How much is v_{0} and how do you evaluate x ? x will go positive or negative ? (get used to use the correct signes).

 

[Thefox14]

Well v₀ is 3 m/s upwards and isn't x just the distance I've stretched? So it would be -.58m right?

 

[Quinzio]

Well v₀ is 3 m/s upwards and isn't x just the distance I've stretched? So it would be -.58m right?

-0.56 m

Ok, so calculate the energy of the system at x = -0.56 m and v0 = 3 m/s

<br /> E = \frac{1}{2}kx^{2} + mgx + \frac{1}{2}mv_{0}^{2}<br />

The same energy must be found at x = -0.28 m

<br /> E = \frac{1}{2}kx^{2} + mgx + \frac{1}{2}mv_{f}^{2}<br />

Solve for v_f

 

[Thefox14]

Ah, that makes sense now. Thanks for the explanation!