Conservation: Mass Dropped onto a Spring, Find the Compression

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Ep=mgh, Ep=1/2k(x)^2

First I wanted to find the kinetic energy the mass had when it hit the spring (converted from the potential Energy it had) thus

Ek=mgh=9.8*2.6*3.5=89.18

Now I know as this Ek changes to 0 the potential energy of the spring as its being compressed will be at its maximum so,

Ek=Ep spring=0.5k(x)^2

Finding the amount its compressed (x), x=sqrt(2*Ek/k)=sqrt(2*89.18/512)=0.590

But the answer is suppose to be 0.452.

 

[TSny]

Ek=mgh=9.8*2.6*3.5=89.18

This does not give you the kinetic energy of the block at the instant it begins to compress the spring because the block doesn't fall a height h in getting to the spring.

Now I know as this Ek changes to 0 the potential energy of the spring as its being compressed will be at its maximum so,

Ek=Ep spring=0.5k(x)^2

As the block compresses the spring, there is a change in gravitational potential energy as well as spring potential energy. Also, note that the final potential energy of the spring is not .5kx². The problem statement says that x is the final length of the spring. But the potential energy of the spring depends on the amount of compression of the spring; that is, on the change in length of the spring.

You do not need to worry about the kinetic energy of the block at the instant the block starts to compress the spring. You just need to compare the initial total energy of the system (at the instant the block is dropped) with the final total energy of the system (at the instant the spring is maximally compressed).

 

[JoeyBob]

You do not need to worry about the kinetic energy of the block at the instant the block starts to compress the spring. You just need to compare the initial total energy of the system (at the instant the block is dropped) with the final total energy of the system (at the instant the spring is maximally compressed).

Wouldn't this be the same though? Total energy of the system at the start is mgh. Total Energy at end would be 1/2kx^2, So mgh=0.5kx^2

 

[haruspex]

Wouldn't this be the same though? Total energy of the system at the start is mgh. Total Energy at end would be 1/2kx^2, So mgh=0.5kx^2

Yes, if you define h suitably. Consider the changes in GPE and EPE between initial and final states.

 

[JoeyBob]

Yes, if you define h suitably. Consider the changes in GPE and EPE between initial and final states.

I've also tried adding the compression to h, using mg(h+x)=0.5kx^2, but this also gives me the wrong answer.

 

[haruspex]

I've also tried adding the compression to h, using mg(h+x)=0.5kx^2, but this also gives me the wrong answer.

I assume your x is the same as the x₀-x in the question.
What do you get numerically?

 

[JoeyBob]

I assume your x is the same as the x₀-x in the question.
What do you get numerically?

Yeah x as the compression.

I get 0.642 and -0.543, neither are right. Also checked my work with Symbolab.

https://www.symbolab.com/solver/algebra-calculator/2.6\cdot9.8\left(3.5+x\right)=0.5\cdot512\left(x^{2}\right)

 

[TSny]

Yeah x as the compression.

I get 0.642 and -0.543, neither are right. Also checked my work with Symbolab.

https://www.symbolab.com/solver/algebra-calculator/2.6\cdot9.8\left(3.5+x\right)=0.5\cdot512\left(x^{2}\right)

In your link you have the equation

If $x$ is the amount that the spring is compressed, then the total distance the mass falls is not $3.5 + x$. The distance will involve $x_0$. A sketch will help.

 

[JoeyBob]

In your link you have the equation

View attachment 274369

If $x$ is the amount that the spring is compressed, then the total distance the mass falls is not $3.5 + x$. The distance will involve $x_0$. A sketch will help.

Wait I am confused here. Are you saying the height is to the bottom of the spring? So (3.6-1.9+x)?

But I thought the height was to the top of the spring, so the compression would just be added on to it (3.6+x)?

 

[TSny]

Wait I am confused here. Are you saying the height is to the bottom of the spring? So (3.6-1.9+x)?

Note that the problem states that $h$ is the height of the block above the Earth's surface.
If you are saying that (3.6-1.9+x) is the total distance that the block falls, then I agree.