Who knew finding Velocity could take so much Work and Energy?

[Byrnkastal]

Hey everyone, first post in here, I'll try to keep it as appropriate as I can.
To put it simply, I'm stumped. The vague setup is a vertical ideal spring attempting to push up a mass against gravity. We have everything we need to get the force, but... I can't calculate the direct transference from acceleration into velocity. Let's see if i can show you what i mean.

Homework Statement

A vertical, ideal spring whose spring constant is 875 N/m is attached to a table and compressed down by 0.160m.
Part a) What upward speed (velocity) can it give to a .380 kg ball when released?
Part b) How high above its original position (in the compressed state) will the ball fly?

Homework Equations

Force=Mass*Acceleration
V[Velocity]= d[Distance, in our case meters] / t [time, in our case seconds]
V= a[acceleration] * t [time] (note: this is specifically for this problem, as there is no initial velocity)
Remember Gravity acting as a force acting against the upward forces (9.8m/s^2)
Force[spring] = K[spring constant]*d[distance from equilibrium, in our case the compression]
We recently were working with Kinetic and Potential energy so i'll include these equations as well...
W[Work] = F[Force]*D[Displacement]
E<k> [Kinetic Energy]= 1/2 M [Mass]* V [Velocity]^2
W<total> [Total Work] = E<k> + E<p> (check me on this in particular)
E<p> [Potential Energy, Specifically for a Spring] = 1/2 K (spring constant) * d [spring displaced due to compression/stretch] ^2 (Credit to Doc Al for catching me, quick search through notes and google and we get this equation, which kinda stands to reason, based off our E<k> equation)

The Attempt at a Solution

I'm frustrated to no end by this particular application. I'm able to find the force applied upwards by the spring, including the counterplay by gravity and mass, but whenever i attempt to find velocity, i can only pull out acceleration- and without a time that the force is applied over, I can't find that velocity. I suppose it could be complicated and messy if i wanted to take the distance compressed, plug it into a few formulas that would give me the time it would take to pass that spot (after passing equilibrium it would stop applying force) but that would require me to incorporate a shifting application of force as well- after all, the spring force is dependent on the distance compressed- the closer to equilibrium it gets, the less actual force it would apply. I feel like this is a problem that I'm simply approaching in the wrong way. Help me, scientificatious people of the internets! T~T

 

[Doc Al]

We recently were working with Kinetic and Potential energy so i'll include these equations as well...
W[Work] = F[Force]*D[Displacement]
E<k> [Kinetic Energy]= 1/2 M [Mass]* V [Velocity]^2
W<total> [Total Work] = E<k> + E<p> (check me on this in particular)

You left out a really important one: What's the elastic potential energy stored in a compressed spring? (Look it up!)

 

[VictorH]

You may be looking at this problem harder than it has to be.

First of all, consider the force applied to the ball. We have a constant weight pushing down vs an elastic, changing, force pulling up. Now, to do this problem following your approach, we would have to determine the total force on the ball as long as it is in contact with the spring. Then, from there, calculate the acceleration and plug this one in your kinematic equations. Kind of long and messy, don't you think?

But, there is another, more manageable, way. Think about this: are all the forces acting on your ball conservative? If so, how does this affect the total work done on your system? How does it affect the kinetic and potential energy of your system?

 

[Rellek]

Remember energy is conserved!

If there's no initial velocity, then the only energy in the system is the initial spring energy. So, how would you model that?

After the spring is released, at what point of the upward descent would your velocity be at a maximum? And don't forget to account for the change in gravitational potential as it moves up!

For the part B, remember that once again the ball would no longer be moving at that top point.

 

[Byrnkastal]

Alright everyone, Since i dislike questions that don't actually post answers or the final solutions, I'm going to put up some of my work. Thanks to help from everyone who posted and from a fellow student on my end, we isolated what was wrong.

The idea for this problem is that (like replies seemed to already know) this is a work-energy problem. For these types of problems, we only really need to remember that the sum energy in the system will be conserved. What isn't kinetic will be potential, and what isn't potential will be kinetic. so when Potential energy is maximum (the energy stored in the spring), kinetic energy will be zero. Vice versa, there won't be any potential energy the instant of/after launch, and the total energy will be equal. Based off that and the equations E<p>=1/2kx^2 (where x is the displacement by the spring), E<k>= 1/2 mv^2, and E<p>=Mass*Gravity*Height, we equate energy stored by the spring to plug in for E<k>, find our velocity for part a. then plug in for MGH, remembering to add .160m, since we're calculating from the compressed point, rather than where it was actually released from all non-gravity forces. Thank you everyone for your help. I appreciate it.
V= 7.47 m/s
H=3.01m

 

[Rellek]

Alright everyone, Since i dislike questions that don't actually post answers or the final solutions, I'm going to put up some of my work. Thanks to help from everyone who posted and from a fellow student on my end, we isolated what was wrong.

The idea for this problem is that (like replies seemed to already know) this is a work-energy problem. For these types of problems, we only really need to remember that the sum energy in the system will be conserved. What isn't kinetic will be potential, and what isn't potential will be kinetic. so when Potential energy is maximum (the energy stored in the spring), kinetic energy will be zero. Vice versa, there won't be any potential energy the instant of/after launch, and the total energy will be equal. Based off that and the equations E<p>=1/2kx^2 (where x is the displacement by the spring), E<k>= 1/2 mv^2, and E<p>=Mass*Gravity*Height, we equate energy stored by the spring to plug in for E<k>, find our velocity for part a. then plug in for MGH, remembering to add .160m, since we're calculating from the compressed point, rather than where it was actually released from all non-gravity forces. Thank you everyone for your help. I appreciate it.
V= 7.47 m/s
H=3.01m

Hey! Your velocity is correct, but it looks like you made a mistake on your height. It looks like you forgot to account for the energy stored in the spring again on the upward ascent!

(If is was pulled down only .16 meters from the resting length of the spring, how could it go more than .16 meters above the resting length?)