# Spring Question Elastic Potential and Kinetic Energy

• chubbyorphan
In summary, a spring with a force constant of 225N/m is compressed 12.0cm from equilibrium by a 1.5kg box. The force exerted on the spring is 27N and the work done on the spring is 1.62J. The maximum speed of the box when released is 1.5m/s. The law of conservation of energy states that the total energy of the system, consisting of elastic potential energy and kinetic energy, remains constant throughout the motion.
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)

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

## The Attempt at a Solution

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!

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.

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

chubbyorphan said:
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.

## 1. What is the spring constant?

The spring constant, denoted by k, is a measure of the stiffness of a spring. It determines how much force is needed to stretch or compress a spring by a certain distance.

## 2. How does the spring constant affect elastic potential energy?

The spring constant directly affects the amount of elastic potential energy stored in a spring. A higher spring constant means a stiffer spring, which requires more force to stretch or compress, resulting in a higher amount of elastic potential energy stored.

## 3. What is the relationship between the spring constant and the frequency of a spring?

The spring constant is inversely proportional to the frequency of a spring. This means that as the spring constant increases, the frequency of the spring decreases, and vice versa.

## 4. How does the mass of an object on a spring affect its elastic potential energy?

The mass of an object on a spring does not directly affect the amount of elastic potential energy stored. However, it does affect the amplitude of the spring's oscillations, which in turn affects the total energy of the system.

## 5. What is the difference between elastic potential energy and kinetic energy in a spring system?

Elastic potential energy is the energy stored in a spring when it is stretched or compressed, while kinetic energy is the energy of motion of an object. In a spring system, the elastic potential energy is converted into kinetic energy as the spring oscillates back and forth.

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