Anyone good in physics? Well I've got a question just for YOU! Please help :)

**audreyaussie** · 11-12-2010 07:10 AM

Two identical mass crates are pushed across the floor by two students. Crate A initially has a speed of 2.0 m/s, while crate B is initially moving with a speed of 6.0 m/s. Explain the difference in the amount of work required to bring the two crates to a stop.

**DrunkenArse** · 11-12-2010 07:33 AM

What definitions of work do you have available to you? There's at least three that you might be using and one of them makes this super easy.

**audreyaussie** · 11-12-2010 07:55 AM

Aha! I think I've got it! I need to calculate their kinetic energy right? o.o

**DrunkenArse** · 11-12-2010 07:59 AM

That's the main point yes. You know that their kinetic energy will end up being 0. So it's just a matter of what their kinetic energy is to begin with. That tells you the work to stop one crate. Then there is the matter of explaining the difference between the two levels of work.

**Invader** · 11-12-2010 09:39 AM

The point of the problem is to test your understanding of how kinetic energy varies with velocity. Crate B has 3 times the velocity of crate A. How much more kinetic energy does crate B have than crate A? That's going to be the difference in work required to bring each crate to a stop.

**DrunkenArse** · 11-12-2010 12:09 PM

Originally Posted by Invader

The point of the problem is to test your understanding of how kinetic energy varies with velocity. Crate B has 3 times the velocity of crate A. How much more kinetic energy does crate B have than crate A? That's going to be the difference in work required to bring each crate to a stop.

Hey, back off! Can't you read the title? The question was just for ME! Quit Invading my thread dude. Oh....wait a minute.

**Philosopher8659** · 11-12-2010 05:13 PM

I would say it is the same for each, simply stop pushing. I do not recall work being defined by the removal of effort.

This seems to be one of those questions for people who think too much but accomplish too little with it.

**DrunkenArse** · 11-12-2010 09:52 PM

Originally Posted by Philosopher8659

I would say it is the same for each, simply stop pushing. I do not recall work being defined by the removal of effort.

This seems to be one of those questions for people who think too much but accomplish too little with it.

I'm sorry, haven't you been banned from helping people with technical homework yet? run along now and go square a circle. And unlike my last comment, I'm not joking.

**Philosopher8659** · 11-12-2010 10:14 PM

Pout.

**Invader** · 11-12-2010 10:44 PM

Originally Posted by Philosopher8659

I would say it is the same for each, simply stop pushing. I do not recall work being defined by the removal of effort.

Work is defined as the amount of energy transfered to a system through forces that act on that system in some distance. Being that all bodies in motion will remain in the same, uniform motion unless external forces are enacted upon them, what do you think causes, say, a crate moving over a flat surface to suddenly slow down? The magical phenomenon at work here (no pun intended) is called friction. This is why you can "simply stop pushing". There is still a force being exerted against the motion of the crate in some distance. Energy is being transferred as kinetic energy to the Earth, and as thermal energy to the contact materials where the crate and ground touch. Differences in velocity are exponentially proportional to the kinetic energy of the system, and differences in mass are directly proportional.

The purpose of the question is to build a fundamental understanding of velocity's relationship to kinetic energy. That is all.

Originally Posted by PhilosopherStoned

Can't you read the title? The question was just for ME!

I see wut you implied thar.

**Philosopher8659** · 11-12-2010 11:01 PM

Originally Posted by Invader

Work is defined as the amount of energy transfered to a system through forces that act on that system in some distance. Being that all bodies in motion will remain in the same, uniform motion unless external forces are enacted upon them, what do you think causes, say, a crate moving over a flat surface to suddenly slow down? The magical phenomenon at work here (no pun intended) is called friction. This is why you can "simply stop pushing". There is still a force being exerted against the motion of the crate in some distance. Energy is being transferred as kinetic energy to the Earth, and as thermal energy to the contact materials where the crate and ground touch. Differences in velocity are exponentially proportional to the kinetic energy of the system, and differences in mass are directly proportional.

The purpose of the question is to build a fundamental understanding of velocity's relationship to kinetic energy. That is all.

I think you need to read the physics book again.

**Invader** · 11-12-2010 11:15 PM

Please point out the specific statements you disagree with and post the corrections to them. Humor me.

The mathematical definition for work, W = F·d = ½m(v₂-v₁)², is not contrary to anything I've said.

**Philosopher8659** · 11-12-2010 11:17 PM

Is work the same as the removal of velocity? Work is always positive.

**DrunkenArse** · 11-12-2010 11:20 PM

Originally Posted by Invader

I see wut you implied thar.

That the title of the thread promised me that because I'm good at physics the question was just for me and made no mention of other people that might be good a physics (or squaring circles)?

Originally Posted by Philosopher8659

This seems to be one of those questions for people who think too much but accomplish too little with it.

Yes. Putting shit in space, airplanes, skyscrapers, cellphones, computers, etc. is "too little". You've made your "point", now please run along.

Originally Posted by Philosopher8659

I think you need to read the physics book again.

I think you need to read the physics book to begin with. And FYI, I'm not talking about Aristotle.

Originally Posted by Philosopher8659

Is work the same as the removal of velocity? Work is always positive.

As defined with Invader's typo, yes it is always positive (so long as v is real). Unfortunately for your assertion, it's a typo.

W = ½m(v₂²-v₁²)

As you can see, this can clearly be negative. It is equivalent to a change in kinetic energy.

**Invader** · 11-12-2010 11:31 PM

Originally Posted by PhilosopherStoned

That the title of the thread promised me that because I'm good at physics the question was just for me and made no mention of other people that might be good a physics (or squaring circles)?

Precisely my good fellow!

Originally Posted by Philosopher8659

Is work the same as the removal of velocity? Work is always positive.

Check it out, we'll define our terms:

Work is the transfer of energy and is expressed in SI units as "Joules".
Velocity is a vector quantity expressed, in SI units, as meters per second, or m/s. It is the distance being covered per second in some direction.

Now, if you asked me again whether or not work is the same as removal of velocity (removal of velocity is change in velocity, dv), what do you think my answer would be? The units they are measured in are not even the same.

Also, work can be negative if energy is being removed from the system in question. Elementary physics, my dear.