Spring Question Elastic Potential and Kinetic Energy

[chubbyorphan]

Homework Statement

Hi forum. I have a spring type question for anyone willing to help!
k = 225N/m
m = 1.5kg
x = 12.0cm
Friction is not a variable

A spring (with a force constant of 225N/m) is resting on a horizontal(frictionless) surface and is mounted on a wall. A 1.5kg box is pushed against the spring compressing it 12.0cm from equilibrium. When released, the spring pushes the box.

a)how much force must be exerted on the spring to compress it 12.0 cm?
b)how much work is done on the spring to compress it 12.0 cm?
c)how much elastic potential energy is stored in the spring it is compressed?
d) Once released, what maximum speed will the box attain?

**I think I got parts a)b)c) if someone wanted to check it anyway though that'd be really cool, but I definitely need help with part d)

Homework Equations

a)F = kx
b)W = (1/2)kx^2

The Attempt at a Solution

here are my answers:

a)F = 27 N
b)E¬e = 1.62 J
c) same as b)
d)We know that maximum speed occurs when Ek is at a maximum and therefore Ee is at a minimum. This point occurs where the spring naturally rests. Therefore, since the spring is compressed 0.120 meters, this point occurs 0.120 meters away upon being released.

Einitial = Efinal
Ee initial = Ek final
1.62 J = (1/2)mv^2
1.62 J = (1/2)(1.5kg)v^2
1.62 J = (0.75kg)v^2
v^2 = (1.62 J)/(0.75kg)
v^2 = (1.62 J)/(0.75kg)
v^2 = 2.16m^2/s^2
v = 1.469693846
After rounding:
v = 1.5m/s
Therefore, the speed of the 1.5kg box as it passes the point 12.0cm or 0.120m from its point of compression will be 1.5m/s

**Is this right^?

Also.. If I wanted to show:
Ee initial +Ek initial = Ee final + Ek final
is this right:
[(1/2)(225N/m)(0.120m)^2][(1/2)(1.5kg)(0)^2] = [(1/2)(225N/m)(0)^2][(1/2)(1.5kg)(v)^2]

Thanks so much in advance to whoever can help me!

 

[SadScholar]

You got it! The block has its maximum velocity when all of its potential energy has been converted into kinetic energy. That is to say that
$\frac{1}{2}kx^2=\frac{1}{2}mv^2$.
Your answers look right to me. You also wanted to know if $E_{elastic}+E_{kinetic}$ changes. This is the total energy of this system, and if the system is closed, such that there is no friction or outside tampering, it must always be the same at any point in time. That is the law of conservation of energy.

 

[chubbyorphan]

Your answers look right to me. You also wanted to know if Eelastic+Ekinetic changes. This is the total energy of this system, and if the system is closed, such that there is no friction or outside tampering, it must always be the same at any point in time. That is the law of conservation of energy.

so does this mean that as the spring passes equilibrium with a speed of 1.5m/s, all of its energy is in the form of kinetic energy at 1.62 J?
Since we started with 1.62 J of solely Elastic potential energy

Thanks so much for your help :D

 

[SadScholar]

so does this mean that as the spring passes equilibrium with a speed of 1.5m/s, all of its energy is in the form of kinetic energy at 1.62 J?

Yeah, you got it.

 

[chubbyorphan]

shweeeeeeet :D thanks SadScholar!
if you got a few minutes I got another thread going about momentum and force! If not its all good you've already helped me a ton!

https://www.physicsforums.com/showthread.php?t=588441