# Conservation of energy - find speed of block

1. Mar 5, 2010

### man_in_motion

1. The problem statement, all variables and given/known data
A 2.4kg block is dropped onto a spring from hieght of 5m. When the block is momentarily at rest, the spring is compressed 25cm. Find the speed of the block when the compression of the spring is 15cm.

2. Relevant equations
$$K=\frac{1}{2}mv^2$$
$$U_g=mgh$$
$$\Delta U = \Delta K$$

3. The attempt at a solution
$$U_o=mgh=2.4kg(9.81m/s^2)(5m)=117.72J$$
since all the potential energy gets converted into kinetic energy
$$117.72J=\frac{1}{2}mv^2$$ then I solve for v and plug the numbers in. I get the right answer but I know I'm doing it wrong because I'm not sure how to take the spring into acount, especially when the springs constant isn't given.

2. Mar 5, 2010

### tiny-tim

Hi man_in_motion!
No, you probably are doing it right.

You don't need to use k, because mgh = 1/2 k 252, and the other PE is 1/2 k 152, so the ks cancel.

3. Mar 5, 2010

### man_in_motion

thanks tiny-tim, though I'm still not entirley convinced.
consider:
$$K_f-K_o = K_f = U_f-U_o=U_f_g+U_f_s-U_g_o=(mg(0.25m)+\frac{1}{2} \cdot \frac{mg}{0.25m}0.15^2m)-mg(5m)=\frac{1}{2}mv^2$$

Then I solve for v and I get
$$v=\sqrt{\frac{2(mg(0.25)+\frac{1}{2} \cdot \frac{mg}{0.25} \cdot 0.15 - mg(5))}{m}}$$
granted now I no longer get 10m/s, but the answer key could be wrong. what do you think?

4. Mar 6, 2010

### tiny-tim

Hi man_in_motion!

(just got up :zzz: …)

Sorry, I don't understand your first line.

5. Mar 6, 2010

### man_in_motion

Energy is conserved
$$K_f+U_f=K_o+U_o$$
$$K_0$$ goes to 0 and rewriting the equation we get $$K_f=U_o-U_f=\Delta U$$
Initially there is only potential energy due to gravity, when the block is in the air
$$U_o_g=mgy=2.4kg(9.81m/s^2)5m$$
There are 2 final potential energies: one for the gravity and one for the spring
$$U_f_g=mgy=2.4kg(9.81m/s^2)(-0.15m)$$
$$U_f_s=\frac{1}{2}kx^2=\frac{1}{2}k(0.15m)^2$$
We don't know k (spring constant) so we need to figure it out using Newton's 2nd law when the block is momentarily at rest

$$\Sigma F_y=ma_y=0=mg-k(0.25)=(2.4kg)(9.81m/s^2)-k(0.25m)=0$$
solving for k I get
$$k=\frac{(2.4kg)(9.81m/s^2)}{(0.25m)}$$

Now we know
$$U_f_s=\frac{1}{2}\frac{mg}{0.25} \cdot (0.15m)^2$$

Plugging things into the first equation

$$K_f=U_f-U_o=mg(0.25m)+\frac{1}{2}\frac{mg}{0.25} \cdot (0.15m)^2-mg(5m) = \frac{1}{2}mv^2$$

sovling for v from the right side of the equation gives
$$v=\sqrt{\frac{2(mg(0.25)+\frac{1}{2}(0.15m)^2-mg(5m))}{m}$$
but this gives me the wrong answer

If I have m following a constant (namley 0.25, 0.15 or 5) it stands for meters (the unit) other wise it's for mass of the block.

Last edited: Mar 6, 2010