What Is the Speed of a Block Launched by a Compressed Spring?

[MAC5494]

Homework Statement

So we have a test this Friday over several topics in modern mechanics. Our professor gave us last year's test to use as a study guide, but hasn't posted the solutions yet. I'm not sure I'm doing this one right so I was hoping someone could check my work! I feel like my number is a bit big.



A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The
spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3
m above its starting position, what is its speed? Be sure to specify what objects are in your system and what
objects are in the surroundings. Show all your work, starting from fundamental principles and/or definitions

system: block, spring, gravity
surrounding: none

U_s = Potential Energy of Spring
U_g = Potential Energy of Gravity
k_s = spring stiffness
s_i & s_f = stretch

Homework Equations

E_f = E_i + W

U_sf + U_gf + K_f = U_si + U_gi + [STRIKE]K_i[/STRIKE] + [STRIKE]W[/STRIKE]

K_f = U_si - U_sf + U_gi - U_gf

.5mv_f² = .5k_s(s_i - s_f)² + mg(y_i - y_f)

v_f² = (ks(s_i - s_f)²)/m + 2g(y_i - y_f)

v_f = sqrt((ks(s_i - s_f)²)/m + 2g(y_i - y_f))

The Attempt at a Solution

I might have messed up the stretch, I'm currently looking over it.

I got v_f = 44.038 m/s

 

[Hollumber]

Seems like you need to double check your s's and y's. Are these potential energies(Us and Ug) supposed to add or subtract? Did you use the correct signs for your s's and y's?

 

[MAC5494]

I used the quantity -.2 for both s_i and y_i since the spring is compressed. I'm not sure what you are saying. U_sf and U_gf should be subtracting.

 

[Hollumber]

I'm not saying what you did is wrong, but to make sure you are consistent with your signs. What values did you use for delta s and delta y?

 

[MAC5494]

delta s and delta y = -1.5

The stretch became +2.25 because it was squared

The thing is though that it is si-sf and yi-yf, so it is different I guess you could say.

 

[Hollumber]

Hmm, I think you can assume all the energy from the spring transfers into the object such that the spring stops decompressing at its equilibrium length. So delta s is just the distance it is compressed, and delta y is how high the block is from where it started.

 

[MAC5494]

Okay I see what you mean. After the spring is released, the block flies off the spring so U_sf is actually just 0, right?

For that, I got Vf = 2.366 m/s which does sound much more realistic haha.

 

[Hollumber]

I think that's around the value I got as well (don't remember, but it's reasonable). Did you use (-)1.3 for delta y? If not, then your velocity is probably a little low.

 

[MAC5494]

No, I used -1.5, because when the block is at 1.3, it is 1.5m above the point it started from.

 

[nasu]

The problem says clearly that is 1.3 m above the starting position. Not 1.5.

 

[MAC5494]

Yeah you are right. Sorry I haven't slept in a while, so I'm kind of delusional.