Work: Dropped vs. Tossed

[pgoyal]

what up gangsters...here's a concept that seems to have my head stumped

lets say your chillin like a villin on top of a building 10m high and drop a ball 1kg. cool no problem.
- the work is simply change in PE = (mgh) which is - from your perspective and + from gravity's perspective. agree?

now, lets say you drop the muthafuken ball like dr.dre drops beats.
case 1. you throw the ball vertically down with a force of 10N. unlike sliding boxes across a table, etc this is a "one time" force you apply to the ball when tossing it.

case 2. you throw the ball vertically down with a velocity of 1m/s (i am not sure if this is even possible...more like you would have to throw the ball with an acceleration of 1 m/s, no?)

in either case what is the work now from the perspective of you and from the perspective of gravity? is it still mgh? i dont think so.

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[mehc012]

what up gangsters...here's a concept that seems to have my head stumped

lets say your chillin like a villin on top of a building 10m high and drop a ball 1kg. cool no problem.
- the work is simply change in PE = (mgh) which is - from your perspective and + from gravity's perspective. agree?

now, lets say you drop the muthafuken ball like dr.dre drops beats.
case 1. you throw the ball vertically down with a force of 10N. unlike sliding boxes across a table, etc this is a "one time" force you apply to the ball when tossing it.

case 2. you throw the ball vertically down with a velocity of 1m/s (i am not sure if this is even possible...more like you would have to throw the ball with an acceleration of 1 m/s, no?)

in either case what is the work now from the perspective of you and from the perspective of gravity? is it still mgh? i dont think so.

Gravity is a conservative force, which means that the path between two states is irrelevant when calculating the work done by gravity to convert between the two. You could take the ball, toss it to your bud on the ground, who would then launch it from a catapult, sending it into orbit, and then collect it after the orbit deteriorated and it plummeted to the ground (assuming it didn't burn up in the atmosphere or anything), and as long as your initial height and your final height were the same, the net work done by gravity would be the same!

The speed at which the ball moves doesn't change the work done by gravity because the same force is being applied by gravity over the same net distance. The same amount of gravitational potential energy is lost no matter the speed of the ball.

What DOES change in your two situations is the kinetic energy/momentum of the ball. This is because there IS additional work done on the ball...it's just not gravitational work. Your arm did work on the ball (though exactly how much would be hard to determine, as most people don't 'apply 10N of force for 0.5m' during a throw), increasing the kinetic energy beyond what gravity would have. Likewise, your throw could be described as providing an impulse (F·t) to the ball and increasing its momentum. However, NONE of these affect anything from gravity's perspective, as you put it (though that's not the clearest phrasing, so correct me if I'm approaching this the wrong way.)

 

[mehc012]

Ugh, I had a huge edit in there which explained net work and conservative forces better, but the clock struck 2AM as I posted, and SDN turned into a pumpkin
Sorry...maybe I'll type it back up when it's not 4AM and I don't have 80 bajillion chapters of Anatomy to cover by 8

 

[pgoyal]

Gravity is a conservative force, which means that the path between two states is irrelevant when calculating the work done by gravity to convert between the two. You could take the ball, toss it to your bud on the ground, who would then launch it from a catapult, sending it into orbit, and then collect it after the orbit deteriorated and it plummeted to the ground (assuming it didn't burn up in the atmosphere or anything), and as long as your initial height and your final height were the same, the net work done by gravity would be the same!

The speed at which the ball moves doesn't change the work done by gravity because the same force is being applied by gravity over the same net distance. The same amount of gravitational potential energy is lost no matter the speed of the ball.

What DOES change in your two situations is the kinetic energy/momentum of the ball. This is because there IS additional work done on the ball...it's just not gravitational work. Your arm did work on the ball (though exactly how much would be hard to determine, as most people don't 'apply 10N of force for 0.5m' during a throw), increasing the kinetic energy beyond what gravity would have. Likewise, your throw could be described as providing an impulse (F·t) to the ball and increasing its momentum. However, NONE of these affect anything from gravity's perspective, as you put it (though that's not the clearest phrasing, so correct me if I'm approaching this the wrong way.)

good, i'm happy that someone else thought the same as myself.

 

[mehc012]

good, i'm happy that someone else thought the same as myself.

Wait, now I'm confused!

 

[pgoyal]

bump! can some1 provide an alternative explanation?

 

[mehc012]

bump! can some1 provide an alternative explanation?

Why don't you explain which part is specifically confusing you, and we'll see if we can phrase it better for you.

 

[pgoyal]

well, for case i & ii could you simply calculate the KE (assuming V is given) right before it touches the ground? if so, would the work done from your perspective be this KE or would it be (KE - mgh).

 

[FeinMS]

Work done by the gravity is the same for both cases.
I. The ball is simply falling down. Let's say height h, mass m. Work done by gravity is simply mgh.
II. The ball travels up and down. Let's assume same parameters and add d (distance that it travels up) First it goes up, goes down to the original height, go further down to the ground. Work done by gravity is (-mgd)+mgd+mgh=mgh

 

[mehc012]

well, for case i & ii could you simply calculate the KE (assuming V is given) right before it touches the ground? if so, would the work done from your perspective be this KE or would it be (KE - mgh).

It would be KE -mgh.

I think what's confusing you is that you're associating the work done by gravity directly with the acceleration of gravity. The two are very closely related, because both result from the constant FORCE applied by gravity, but the work is not really done by the acceleration.

Work done by gravity: F_grav · d

Note that this is the same anytime the height of the object changes by 'd', no matter the path.

acceleration due to gravity: F_grav /m_obj

This is constant over TIME in a free fall, meaning that if an object covers a distance in 2 seconds, it will have accelerated by 20m/s by the time it hits the ground. If you propel it at the ground and it hits after only 1 second, than the total acceleration done by gravity is only 10 m/s. THIS DOES NOT MEAN THAT LESS WORK WAS DONE BY GRAVITY, because work is related to DISTANCE, not time.


To put it in a perspective that we're more used to calculating, consider what happens when you throw something UP: it's fairly easy to recognize that you must do enough work to counteract that of gravity, plus however much is needed to actually provide KE for the object (give it velocity).
That is to say, it's easy to see why W_hand = mgh + KE in that instance; the same is true when propelling an object downwards, only mgh will have a negative value, so W_hand = KE + mg·(-h) = KE - mgh

 

[pgoyal]

no no, i totally understand the gravity part.

I am more focused on the Individual throwing the ball part. normally questions as for work done by gravity...but I can totally imagine the mcat asking instead work done by the person.

so for simply dropping the ball work done by person is 0 right? b/c Fperson = 0. therefore Fperson * d = 0

and for case i/ii like you said work done by person would be KE - mgh.

 

[FeinMS]

no no, i totally understand the gravity part.

I am more focused on the Individual throwing the ball part. normally questions as for work done by gravity...but I can totally imagine the mcat asking instead work done by the person.

so for simply dropping the ball work done by person is 0 right? b/c Fperson = 0. therefore Fperson * d = 0

and for case i/ii like you said work done by person would be KE - mgh.

They don't ask what the work done by the person is because it's 0. Why is it 0? because you're not applying the force continuously over the displacement unlike gravity. For example, when you're pushing a box on a horizontal surface, you're doing work because you keep pushing (keep applying force). But in this case, you're simply dropping it (case I) or giving it some momentum over a negligible distance (case II). Yeah I know case II can be argued upon because you're providing the force (change in momentum over time) for really small distance and that may count as work. Correct me if I'm wrong

 

[mehc012]

They don't ask what the work done by the person is because it's 0. Why is it 0? because you're not applying the force continuously over the displacement unlike gravity. For example, when you're pushing a box on a horizontal surface, you're doing work because you keep pushing (keep applying force). But in this case, you're simply dropping it (case I) or giving it some momentum over a negligible distance (case II). Yeah I know case II can be argued upon because you're providing the force (change in momentum over time) for really small distance and that may count as work. Correct me if I'm wrong

In case 1, you are DEFINITELY not doing work, because a) you are not in contact with the object over any displacement and b) you're not providing a force.

In case 2, you ARE doing work...the main hole in your reasoning being the dismissal of your actions as occurring 'over a negligible distance'. If the force is high enough, the distance need not be very great to do a lot of work on an object...think again of the case where you throw a ball UP, instead of down. Clearly, you are capable of doing enough work over the course of a throw to impart a significant amount of KE on the object. In fact, if you can measure how high you can throw an object, you start to get a sense of how much work you can actually do in that distance! For example, an object starts at rest (KE = 0) and is thrown hard enough (KE = ?) that it travels 10m upwards before its KE again equals 0. In order to decrease the KE from ? to 0, we know that gravity must have done negative work with a magnitude of mgh (or 100·m)...meaning that you must have done that much POSITIVE work to increase its KE to ? in the first place!
The same principles apply when you are throwing an object down, only it's more difficult to measure the change in KE (you'd have to measure velocity directly)...you are exerting a force over a distance, and so you are doing work on the object.

*for an example of large work being done over a VERY short distance, consider a gun...the force is sufficiently massive to compensate for the size

**Also consider that whether or not the distance is 'negligible' depends on the situation...if you're sitting on a 20-ft building and you do a full wind up spike, you're probably in contact with the ball for a good 4ft or so...almost 20% of the height it will travel, being worked on by gravity, after release! Now, if you're on the Empire state building (in a vacuum, to prevent terminal velocity), that's not significant...but that will be reflected in the calculations using W_you = KE - mgh

 

[FeinMS]

In case 1, you are DEFINITELY not doing work, because a) you are not in contact with the object over any displacement and b) you're not providing a force.

In case 2, you ARE doing work...the main hole in your reasoning being the dismissal of your actions as occurring 'over a negligible distance'. If the force is high enough, the distance need not be very great to do a lot of work on an object...think again of the case where you throw a ball UP, instead of down. Clearly, you are capable of doing enough work over the course of a throw to impart a significant amount of KE on the object. In fact, if you can measure how high you can throw an object, you start to get a sense of how much work you can actually do in that distance! For example, an object starts at rest (KE = 0) and is thrown hard enough (KE = ?) that it travels 10m upwards before its KE again equals 0. In order to decrease the KE from ? to 0, we know that gravity must have done negative work with a magnitude of mgh (or 100·m)...meaning that you must have done that much POSITIVE work to increase its KE to ? in the first place!
The same principles apply when you are throwing an object down, only it's more difficult to measure the change in KE (you'd have to measure velocity directly)...you are exerting a force over a distance, and so you are doing work on the object.

*for an example of large work being done over a VERY short distance, consider a gun...the force is sufficiently massive to compensate for the size

**Also consider that whether or not the distance is 'negligible' depends on the situation...if you're sitting on a 20-ft building and you do a full wind up spike, you're probably in contact with the ball for a good 4ft or so...almost 20% of the height it will travel, being worked on by gravity, after release! Now, if you're on the Empire state building (in a vacuum, to prevent terminal velocity), that's not significant...but that will be reflected in the calculations using W_you = KE - mgh

I see it makes sense. So you're saying since the kinetic energy of the ball is changing at the moment of throwing the ball, I am doing work right? Work by me=change in KE. And at the moment when the ball hits the ground, it will have KE from me and KE from potential energy so W by me=KE final - mgh right? Thanks for the correction

 

[mehc012]

I see it makes sense. So you're saying since the kinetic energy of the ball is changing at the moment of throwing the ball, I am doing work right? Work by me=change in KE. And at the moment when the ball hits the ground, it will have KE from me and KE from potential energy so W by me=KE final - mgh right? Thanks for the correction

 

[FeinMS]

You killed me with your brain