Force of Falling Objects at a Height

[Samei]

Homework Statement

: [/B]I'm curious to know why and how one would account for the force of a falling object dropped from a height. If I apply Newton's 2nd Law, force is only dependent on acceleration. So in a straight vertical drop, this acceleration is only gravity. But is it not that a bowling ball dropped from a skyscraper will have more force crashing down into the ground than the same ball dropped from only half a meter of a height?

Here is my problem: If I drop a ball into a spring from a given distance x, the spring will compress. Assume that the ball sticks to the spring. After impact, the spring will eventually have a restoring force to counteract that force. But how much is that force? It is not equal to the weight of the ball, is it? (In this case, the spring is massless. Only mass of the ball (m), spring constant (k), and distance of the ball from spring (x) is given.)

Homework Equations

: [/B]F = m * a, F (spring) = -kx, W = KE2 - KE1, Conservation of Energy, W = Fd

The Attempt at a Solution

: [/B]W = KE2 - KE1
KE2 = mgx by conservation of energy, x is the height
KE1 = 0, because ball is being dropped

W = KE2 = mgx
and F = W/d

Substituting,
F = (mgx)/d
d = x + n, n is the distance the spring is compressed due to the weight of the ball

I am unsure how to solve for n.

F = (mgx)/(x+n) = -kn
I'm stuck here. It looks like a quadratic, but it looks incorrect.

Once I solve for n, I would have -kn, which is the restoring force.

Can someone help me? Thanks in advance!

 

[Qwertywerty]

Firstly , the reason for any and all deformation after collision is pretty much only dependent on momentum , and not force .

Here is my problem: If I drop a ball into a spring from a given distance x, the spring will compress. Assume that the ball sticks to the spring. After impact, the spring will eventually have a restoring force to counteract that force. But how much is that force? It is not equal to the weight of the ball, is it?

It depends . Which instant are we talking about ? If we speak of the instant at which it just touches the spring , it will be zero .

If we speak of the instant at which net force on the object is zero , yes , it will be equal to mg .

Your solution ... I can't understand what you have done .

 

[paisiello2]

But how much is that force? It is not equal to the weight of the ball, is it? (In this case, the spring is massless.

Correct.

Doesn't W = mg(x+n) ?

Also doesn't the spring do work on the mass as well?

 

[Qwertywerty]

Correct.

Doesn't W = mg(x+n) ?

Also doesn't the spring do work on the mass as well?

The solution seems wrong .

 

[Ellispson]

Homework Statement

: [/B]I'm curious to know why and how one would account for the force of a falling object dropped from a height. If I apply Newton's 2nd Law, force is only dependent on acceleration. So in a straight vertical drop, this acceleration is only gravity. But is it not that a bowling ball dropped from a skyscraper will have more force crashing down into the ground than the same ball dropped from only half a meter of a height?

Here is my problem: If I drop a ball into a spring from a given distance x, the spring will compress. Assume that the ball sticks to the spring. After impact, the spring will eventually have a restoring force to counteract that force. But how much is that force? It is not equal to the weight of the ball, is it? (In this case, the spring is massless. Only mass of the ball (m), spring constant (k), and distance of the ball from spring (x) is given.)

Homework Equations

: [/B]F = m * a, F (spring) = -kx, W = KE2 - KE1, Conservation of Energy, W = Fd

The Attempt at a Solution

: [/B]W = KE2 - KE1
KE2 = mgx by conservation of energy, x is the height
KE1 = 0, because ball is being dropped

W = KE2 = mgx
and F = W/d

Substituting,
F = (mgx)/d
d = x + n, n is the distance the spring is compressed due to the weight of the ball

I am unsure how to solve for n.

F = (mgx)/(x+n) = -kn
I'm stuck here. It looks like a quadratic, but it looks incorrect.

Once I solve for n, I would have -kn, which is the restoring force.

Can someone help me? Thanks in advance!

F=W/d only when the force is constant.Is the spring force constant here?

 

[CWatters]

Homework Statement

: [/B]I'm curious to know why and how one would account for the force of a falling object dropped from a height. If I apply Newton's 2nd Law, force is only dependent on acceleration. So in a straight vertical drop, this acceleration is only gravity.

No you are confusing the acceleration and deceleration phase. The "acceleration" that determines the impact force is the deceleration when it hits the ground not the acceleration on the way down (that's constant if you ignore air resistance).

 

[CWatters]

Perhaps continue with the COE theme and google "Energy stored in a spring".

 

[Samei]

Sorry for the delay, but I'm back!

Firstly , the reason for any and all deformation after collision is pretty much only dependent on momentum , and not force .

It depends . Which instant are we talking about ? If we speak of the instant at which it just touches the spring , it will be zero .

If we speak of the instant at which net force on the object is zero , yes , it will be equal to mg .

Your solution ... I can't understand what you have done .

At the instant it touches the spring, then it will be zero. But what about after it compresses the spring? Since it pushes down the spring, I imagine it would gain more force. I wasn't sure of my equations either, but my main thoughts were of the fact that spring force (-kn) is equal to the force applied to it (the spring).

Correct.

Doesn't W = mg(x+n) ?

Also doesn't the spring do work on the mass as well?

Because it compresses the spring, right? So it is x (height) and n (compressed distance). The spring does work through its restoring normal force.

F=W/d only when the force is constant.Is the spring force constant here?

Not sure about this, but I am going to say it acts like a "standard" spring, so spring constant is constant (k).

Perhaps continue with the COE theme and google "Energy stored in a spring".

I just read a post regarding this, with the author summarizing it as the "fall doesn't kill you, but the deceleration (impact) does". I have since attempted to apply his strategy but found that I could not. I could use Impulse, but I am lacking t (time), and possibly other variables.Thanks to everyone who replied. I am still working on it. Right now, I am looking at using Impulse, although I am thinking Work-Energy Theorem should explain this. Am I overcomplicating this problem?

 

[Samei]

I really do appreciate the input, especially since my textbook does not explain this. :)

 

[Ellispson]

Not sure about this, but I am going to say it acts like a "standard" spring, so spring constant is constant (k).

Stage spring constant is definitely constant.But the spring force is proportional to the compression in the spring.This means the force is not constant.Now F=W/d(This is what you've used.)is only valid for a constant force,hence it won't work here.

 

[Samei]

Stage spring constant is definitely constant.But the spring force is proportional to the compression in the spring.This means the force is not constant.Now F=W/d(This is what you've used.)is only valid for a constant force,hence it won't work here.

Oh, ok. That would explain why it did not seem right. So it looks like I'll have to use another set of equations then. I'll post what I'll find, which may take a moment or so. I'm looking at Impulse right now.

 

[Samei]

Alright. So, impulse was not the solution.

It turns out that I only need to use conservation of energy.

Here is my redo with same variables applied: mgx = Total E = KE1
Again, x is the height, m is the mass of the ball, and k is the spring constant as the given. n will be the compression of the spring.

At the instant it touches the spring, KE1 = mgn + 0.5kn^2 + KE2
KE2 = 0, since the ball stops at max compression.
KE1 = mgn + 0.5kn^2
It is a quadratic again. But solvable for n with given information.

Once I obtain this n, I simply plug it into -kn and I will have my restoring force.

I realize that there might be collision (inelastic/elastic) elements here. I just made up this problem to understand some topics so it is a bit flawed. Is the solution right, at least conceptually? I may ask another similar problem again if it helps.

 

[Ellispson]

Alright. So, impulse was not the solution.

It turns out that I only need to use conservation of energy.

Here is my redo with same variables applied: mgx = Total E = KE1
Again, x is the height, m is the mass of the ball, and k is the spring constant as the given. n will be the compression of the spring.

At the instant it touches the spring, KE1 = mgn + 0.5kn^2 + KE2
KE2 = 0, since the ball stops at max compression.
KE1 = mgn + 0.5kn^2
It is a quadratic again. But solvable for n with given information.

Once I obtain this n, I simply plug it into -kn and I will have my restoring force.

I realize that there might be collision (inelastic/elastic) elements here. I just made up this problem to understand some topics so it is a bit flawed. Is the solution right, at least conceptually? I may ask another similar problem again if it helps.

Yes it is correct.

 

[Samei]

Yes it is correct.

Thank you!