Spring Physics Problem Solution: Launching Rocks to Survive on a TV Show

[Char. Limit]

Homework Statement

"The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch high in a tree. Fortunately, they have a spring they can use to launch a rock. The spring constant is 1200 N/m, and they can compress the spring a maximum of 34 cm. All the rocks on the island seem to have a mass of 500 g."

So...

$$k=1200 N/m$$

$$\Delta s = - 0.34 m$$

$$m_r = 0.50 kg$$

Homework Equations

$$F_s=-k\Delta s$$

The Attempt at a Solution

So what I tried was saying that since k=1200 N/m, and s=-.34 m, then F_s=-ks=408 N. So I thought, well, this is the mass of the rock times the acceleration of the rock, right? So I did...

$$F_s=408 N = m_r a_s = 0.5 a_s...a_s=816 m/s^2$$

Then, I tried to apply the kinematics equation...

$$v_f^2=v_0^2 + 2 a x$$

To try to find the speed of the rock when it left the spring. So, I guessed that v_0=0, a=816-9.8=806.2 m/s^2(accounting for gravity), and x=.34 m. I plugged these values into get

$$v_f=23.4$$

Problem is, the homework grader says it's wrong. What did I do wrong?

 

[The legend]

but that spring acceleration is only till it leaves the spring and that is not constant... 'a' by spring changes with time... so think about it again..

 

[Char. Limit]

Oh wow, I can't believe I forgot the question.

Here it is:

"With what speed does the rock leave the spring?"

Hmm... since a isn't constant, I'm guessing that we have to do an integration here, is that right?

...

no, that probably isn't right.

Can we use a conservation of energy theorem here, then?

 

[The legend]

conservation of energy sounds good!

 

[Char. Limit]

Hmm... I tried it, but it didn't work as well as I thought.

So, it should be...

$$\frac{1}{2} m v_f^2 + \frac{1}{2} k s_f^2 = \frac{1}{2} m v_i^2 + \frac{1}{2} k s_i^2$$

So, since v_i and s_f both are 0 (I think), we get this...

$$\frac{1}{2} m v_f^2 = \frac{1}{2} k s_i^2$$

Solving for v_f, we have...

$$v_f=\sqrt{\frac{k s_i^2}{m}}$$

plugging in my values for those, I get v=16.66, which since the question asks for two significant figures, I rounded to v=17. But I still got it wrong for some reason.

 

[The legend]

oh forgot... if u compress the spring then leave it it goes to normal position and continues to some height... so at h=0.34 the s_f won't be 0... include this in your solution..

 

[Char. Limit]

Hmm... I just realized, this is a vertical spring. Should I be taking gravity into account at all?

 

[The legend]

yes you should.. because that's what causes the gravitational potential energy of the ball.

 

[Char. Limit]

OK, so I have a new question. How exactly do I find h for mgh, the gravitational potential energy?

 

[The legend]

consider the compressed place as the reference point...so the potential energy there =0... then at what height the stone leaves the contact with spring is the h for mgh.

 

[Char. Limit]

After including gravity, I got v=16 m/s, which was correct. Thanks for the help!

 

[The legend]

you are welcome!
which grade are you in by the way?

 

[Char. Limit]

College freshman.

 

[The legend]

oh right!