# Elastic potential energy (spring) problem

• newguy_13
In summary, when a 10.0 N force is exerted on a 0.200 kg object attached to a horizontal spring with a spring constant of 77.0 N/m, the spring is compressed to a distance of 2.54m.
newguy_13

## Homework Statement

A 0.200 kg object is attached to a horizontal spring with a spring constant of 77.0 N/m. The other end of the spring is attached to a wall in such a way that it rests on a frictionless horizontal surface. A 10.0 N force is exerted on the spring, causing it to compress

a) How far will the spring be compressed?
b) how much work is done on the spring to compress it?
c) Once the force on the spring is released, the spring and object are set into harmonic motion. What will the maximum speed of the object be? Explain where this speed will occur. (no friction and mass of spring is negligible)
d) What will spring's position be when v=1m/s?

Ee=1/2kx^2

## The Attempt at a Solution

a)
f=kx
x=f/k
x=0.13m
b)
Ee=1/2kx^2
=1/2(77)(0.13)^2
=0.649J
c)
1/2mv^2=1/2kx^2
1/2(0.2)v^2 = 0.649
v=2.54m/s
d)
1/2mv^2 = 1/2kx^2
0.1=1/2(77)x^2
2(0.1)/77=x^2
0.051m=x

Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2

newguy_13 said:
Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2
That's the correct approach: energy is conserved. Your mistake in part d was setting the KE (when v=1m/s) equal to the total energy.

In part c you got lucky since when the KE is max the PE is zero. (Where does that happen?)

collinsmark
newguy_13 said:

## Homework Statement

A 0.200 kg object is attached to a horizontal spring with a spring constant of 77.0 N/m. The other end of the spring is attached to a wall in such a way that it rests on a frictionless horizontal surface. A 10.0 N force is exerted on the spring, causing it to compress

a) How far will the spring be compressed?
b) how much work is done on the spring to compress it?
c) Once the force on the spring is released, the spring and object are set into harmonic motion. What will the maximum speed of the object be? Explain where this speed will occur. (no friction and mass of spring is negligible)
d) What will spring's position be when v=1m/s?

Ee=1/2kx^2

## The Attempt at a Solution

a)
f=kx
x=f/k
x=0.13m
b)
Ee=1/2kx^2
=1/2(77)(0.13)^2
=0.649J
c)
1/2mv^2=1/2kx^2
1/2(0.2)v^2 = 0.649
v=2.54m/s
d)
1/2mv^2 = 1/2kx^2
0.1=1/2(77)x^2
2(0.1)/77=x^2
0.051m=x

Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2

Hello @newguy_13,

Welcome to PF!

a) through c) look good to me with the possible exceptions of minor rounding errors.

Don't forget that c) requires you to specify where the maximum speed occurs. (You forgot to specify the location.)

For d), use your second approach. Your first approach is not correct.

It turns out that you already used your second approach when doing part c), even if you didn't realize it. (At the point of maximum speed, what is the system's potential energy? )

Thanks so much for the help guys!

## 1. What is elastic potential energy?

Elastic potential energy is the energy stored in an object when it is stretched or compressed. It is a type of potential energy that is caused by the deformation of an object, such as a spring, and can be released as kinetic energy when the object returns to its original shape.

## 2. How is elastic potential energy calculated?

The formula for calculating elastic potential energy is E = ½ kx², where E is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

## 3. What is the relationship between elastic potential energy and the spring constant?

The spring constant is a measure of the stiffness of a spring and is directly proportional to the amount of elastic potential energy stored in the spring. A higher spring constant means a stiffer spring and more elastic potential energy stored.

## 4. Can elastic potential energy be negative?

Yes, elastic potential energy can be negative. This occurs when the spring is compressed and the displacement is negative. In this case, the elastic potential energy is also negative, indicating that the spring has stored energy that can be released when it returns to its equilibrium position.

## 5. How is elastic potential energy used in real life?

Elastic potential energy is used in many everyday objects, such as trampolines, bungee cords, and pogo sticks. It is also used in more complex systems, such as in car suspension systems, where the elastic potential energy of springs helps to absorb shocks and vibrations.

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