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[Shortround6]

Now my head hurts in addition to my back.

 

[pbehn]

Now my head hurts in addition to my back.

I think the discussion is working around in circles.

 

[ThomasP]

Hey Mike,

The answer is in the math.

In the example of the couch above, lets say that Shortround pushes the couch from V0 at X0 over a distance D1 in the positive X direction, using a force F1 resulting in a velocity V1. Shortround then moves (instantly) to the opposite side of the couch and pushes in the opposite direction using a force F2 over a distance D2 in the positive X direction until the couch comes to a velocity of V0 at X2. Since the Work (W1) done to accelerate the couch to V1 has to be equal to the Work (W2) done to decelerate the couch to a velocity of V0, the net work is 0 even though the couch is displaced by a distance of D1 + D2.

This is because the energy state of the couch is the same at X2 as it was at X0.

You can substitute any values in the above formula as long as the result is a velocity at X0 that equals the velocity at X2.

 

[Zipper730]

Okay, let's get back to it. I'll put up my chart, and you guys tell me if you like the way it looks, if all the numbers are filled out decently.

 

[Zipper730]

BTW: How do you switch axes in Excel graphs?

 

[wuzak]

I think the mmHg would probably be absolute pressure. It would depend on teh country, though.

 

[wuzak]

Hey Mike,

The answer is in the math.

In the example of the couch above, lets say that Shortround pushes the couch from V0 at X0 over a distance D1 in the positive X direction, using a force F1 resulting in a velocity V1. Shortround then moves (instantly) to the opposite side of the couch and pushes in the opposite direction using a force F2 over a distance D2 in the positive X direction until the couch comes to a velocity of V0 at X2. Since the Work (W1) done to accelerate the couch to V1 has to be equal to the Work (W2) done to decelerate the couch to a velocity of V0, the net work is 0 even though the couch is displaced by a distance of D1 + D2.

This is because the energy state of the couch is the same at X2 as it was at X0.

You can substitute any values in the above formula as long as the result is a velocity at X0 that equals the velocity at X2.

Your logic means there would never be any work done where the couch has the same final velocity and altitude.

Work is simply the force multiplied by the distance. The total work in your example is W = F1.D1 + F2.D2.

W=0 only when F1.D1 = -F2.D2

The thing missing from your calculation is the time between the acceleration phase and the deceleration phase, where the velocity and force remain constant.

Also, in the couch example, friction works against you for acceleration and with you for deceleration. Not often when moving furniture do you have to run to the other side and push to slow it down.

 

[Zipper730]

I think the mmHg would probably be absolute pressure. It would depend on teh country, though.

I'm personally a fan of absolute pressures myself, but the Japanese used gauge pressure in mmHg.

 

[Zipper730]

When it comes to making graphs on Excel. I've figured out how to alter the axes for climb & altitude but when I select altitude and climb-speed, I get some very funny numbers.

The altitude section looks normal but the speed figures are like 0-20. Instead of 151.5 being the lowest climb-speed listed.

 

[mikewint]

The answer is in the math.

It is indeed except that you are equating different types of work. On a flat level friction-less surface W is simply F (force applied) X D (displacement)
In your scenario you are including the force needed to produce an acceleration to a constant velocity. In this case W = Δ KE. Considering that the couch was initially at rest, initial KEi = 0 thus Wt = (0.5)mv^2. On a frictionless surface the couch will continue to move at a constant velocity unless an opposing force is applied. When that opposing force is applied and it is equal and opposite to the initial force then the couch will come to rest. So the Work done to accelerate and the work done to decelerate (net Work) do indeed total zero in this frictionless pretend world but in the real world friction has to be contended with and a constant force must be applied to maintain that constant motion. That "extra" force is in the direction of motion and there has been a change in d (displacement) so that work remains unchanged since W still = F d

 

[pbehn]

It all depends what you want to know, a plough ends up back in the same place it started at after a day ploughing a field, that doesn't mean that no work has been done.

 

[mikewint]

that doesn't mean that no work has been done.

Depends on how you define "WORK". In Physics WORK (one definition) is defined as FORCE x DISPLACEMENT and the displacement must be in the same direction as the force applied AND WORK is a vector quantity so a displacement North followed by an equal displacement South are adding a + number to a - number which totals Zero like it or not. So NO NET Work was done. Positive work was done as the plow went North and Negative Work was done as the plow moved South. So individual quantities of Work were indeed done but the NET TOTAL Work remains zero. By the same token you can hold up a weight until your arm turns black and falls off but NO Displacement means NO Work was done. Push on your house until you collapse in a puddle of sweat. No displacement NO Work.
Consider Post #23. WORK is also defined as WORK = Δ KE or Work(t) = KEi + KEf. IF we live in a frictionless world then object in motion remain in motion. Thus we can apply an instantaneous force to accelerate to a velocity and the object moves at constant velocity with no extra force applied. There has been a change in KE thus Work was done. At some point in time apply a second instantaneous equal and opposite force and the object stops. There has been a second change in KE. BUT that first KE change was positive and the second was negative thus WORK TOTAL = Zero and the weird part is that during the constant velocity phase a displacement occurred BUT BUT BUT even though Work = F d NO WORK was done to produce this d since the F(net) was zero during this constant velocity phase. PHYSICS IS PHUN!!!

 

[pbehn]

Depends on how you define "WORK". In Physics WORK (one definition) is defined as FORCE x DISPLACEMENT and the displacement must be in the same direction as the force applied AND WORK is a vector quantity so a displacement North followed by an equal displacement South are adding a + number to a - number which totals Zero like it or not. So NO NET Work was done. Positive work was done as the plow went North and Negative Work was done as the plow moved South. So individual quantities of Work were indeed done but the NET TOTAL Work remains zero. By the same token you can hold up a weight until your arm turns black and falls off but NO Displacement means NO Work was done. Push on your house until you collapse in a puddle of sweat. No displacement NO Work.
Consider Post #23. WORK is also defined as WORK = Δ KE or Work(t) = KEi + KEf. IF we live in a frictionless world then object in motion remain in motion. Thus we can apply an instantaneous force to accelerate to a velocity and the object moves at constant velocity with no extra force applied. There has been a change in KE thus Work was done. At some point in time apply a second instantaneous equal and opposite force and the object stops. There has been a second change in KE. BUT that first KE change was positive and the second was negative thus WORK TOTAL = Zero and the weird part is that during the constant velocity phase a displacement occurred BUT BUT BUT even though Work = F d NO WORK was done to produce this d since the F(net) was zero during this constant velocity phase. PHYSICS IS PHUN!!!

That is what I said.

 

[Joe Broady]

The problem is I'm not sure what calibrations were used at the time to create the figures. For example 14.70 is commonly seen, it seemed the RAF used 14.696. Technically 14.69595 can be used, but I'm not sure if they entered those numbers, and I'm figuring it might have a slight effect that wasn't in the actual documents.

Do you realize that 14.7 psi, which you probably learned as a schoolboy, is within 0.03% of the standard sea level pressure of 1.01325 bar? That's an error of 1 part in 3600. I think you're agonizing over trifling differences between numbers. Isn't 0.03% accurate enough for what you're doing?

A sign of a numerate person is precision which is sufficient but not excessive.

 

[ThomasP]

Hey Wuzak,

re your post#27: "Your logic means there would never be any work done where the couch has the same final velocity and altitude."

Please reread my post#23. My statement is that there was 0 net work. You can add in additional F3 x D3 (ie friction), F4 x D4 (ie cohesion), F5 x D5 (ie wind resistance), Fn x Dn, etc .... it will not change the value of net work if the velocity at Xn ends up equal to 0.


Hey Mike and Wuzak,

I think there may be a slight misunderstanding here as to what work actually is. Work (physics definition) is defined as doing something to an object or system that changes its energy state.

In a mechanical system this is usually defined as changing the energy state of the object/system by applying a force over a distance (ie F x D) which would change its kinetic energy (KE), which is why W=ΔKE. Since Force is equal to Mass x Acceleration (ie F=M x A), in order to have net work (ie ΔKE) you would have to have accelerated the object/system to a velocity not equal to the start velocity.

In Mike's post#11 he states: "Raising a 20 pound weight to a height of 4 feet above the floor and then lowering it back down does zero total work since the positive lifting work and the negative lowering work cancel." The reason that this true is that the work done by raising the 20 pound weight 4 feet raised the Potential Energy (PE) of the object and lowering the object reduced the Potential Energy of the object by the same amount. In this case W=ΔPE, and since the PE ended up the same after the object was lowered back to its start height there was 0 net work. It should be noted that this would also be true if the object was raised 4 feet, moved sideways 4 feet, and then lowered 4 feet back to the floor, ie ΔPE=0

As I stated in my post#19, work may be described in equivalences, ie mechanical (ΔKE), heat (ΔT), magnetic (ΔG), etc. In none of the equivalences will you find the result of work listed as just feet, which is what you would have to be able to do if displacement without a net change in energy state counted as work. I am not trying to be snarky here but go ahead and try to find any variation of work done, net or otherwise, described as
"= x feet" or "Δft".

 

[wuzak]

[When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement. There are several good examples of work that can be observed in everyday life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced./QUOTE]

Definition and Mathematics of Work

In physics, work is defined as a force causing the movement—or displacement—of an object. In the case of a constant force, work is the scalar product of the force acting on an object and the displacement caused by that force. Though both force and displacement are vector quantities, work has no direction* due to the nature of a scalar product (or dot product) in vector mathematics. This definition is consistent with the proper definition because a constant force integrates to merely the product of the force and distance.

What's the Definition of Work in Physics, and How Do You Calculate It?

* This contradicts my earlier statement of work being a vector quantity.

Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion times the distance through which the force acts.

Work: The Scientific Definition – College Physics

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. Energy is required to do work and the basic SI unit of energy is the joule, the amount of energy required to exert a force of 1 Newton through a distance of 1 meter (1 joule = 1 newton meter).

Work

 

[mikewint]

any variation of work done, net or otherwise, described as
"= x feet" or "Δft".

In Ye Olde Golden Days of Yore Physics problems were worked in one of three coherent systems of measurement. Either the MKS or the CGS or the FPS systems, MKS - Meter Kilogram Second; CGS - Centimeter Gram Second; or the English FPS - Foot Pound (unit of Force) Second.
So the Work formula W = F x d could have 3 answers depending on the system used, i.e. MKS - 20 Newtons x 10 meters = 200 Nm or Newton Meters; CGS - 20 dynes ( 1 g⋅cm/s^2) x 10 centimeters = 200 ergs (1 g⋅cm^2/s^2); and finally FPS - 20 pounds x 10 feet = 200 ft⋅lb.
In the US strangely torque wrenches are calibrated in inch-pounds or foot-pounds(ft-lb) (the torque produced by a one pound weight hanging at the end of a 1 foot long lever)
With the almost universal switch to the SI system today Physics problems are always couched in the MKS system.
Historically the very term "WORK" was introduced by a French mathematician in 1826 and defined as "weight lifted through a height". The usage originated from measuring the output of early steam engines lifting buckets of water from flooded mine shafts. In the SI system the work unit is the JOULE which is defined as one newton of force acting through one meter of displacement. To make things even funner: this would be a Newton-meter but that is also a Torque unit which is also defined as force x distance or N-m. The Joule is also a unit of energy so WORK transfers ENERGY. Joule also measures HEAT energy since a Joule of Heat is produced by a current of one ampere passing through a resistance of one Ohm for one second.
WORK and KINETIC ENERGY are also related in that the WORK done by all forces acting on a particle equals the change in the particles KINETIC ENERGY.
Thus WORK can change the POTENTIAL ENERGY of a mechanical system; the THERMAL ENERGY of a Thermal system; the ELECTRICAL ENERGY in an Electrical system. In effect WORK transfers ENERGY from one place to another or from one form to another

 

[AMCKen]

As mikewint said "In passing: one standard atmosphere will support a column of Mercury 760 mm tall thus 1 ATM = 760 mm of Hg. Pre SI, mm of Hg were called TORR "

TORR for Torricelli, the fellow who invented the barometer.
Evangelista Torricelli - Wikipedia