/0's physics phread

[Mao]

The force on the car is opposite the displacement of the car, the work done by the truck on the car is negative (that is, the car did work on the truck).

[/0]

Hey does anyone know what the units for are? Are they meant to be or ?

[evaporade]

hmmm

If a car is moving at 40m/s and a truck collides with it, stopping the car, would you say the work the truck does on the car is negative or positive? The answers say it's positive but how could that be, seeing as the car's change in kinetic energy is negative?

When you say work is done ON an object what do you mean?
e.g. If a truck hits a car, stopping it, is it the work exerted by the truck, or the work done [/i]to the system which is the car?

From the moment the truck comes in contact with the car until it stops, the truck moves forwards and the force exerted by the truck on the car is also in the forward direction. So the work done by the truck is positive because the displacement of the truck (and the car) and the force of the truck are in the same direction.

[evaporade]

In fact the car and the truck question is quite complicated. The answer depends on the mass of the car, mass of the truck, velocity of the truck and whether friction is involved.

If there is no friction involved, the car stops when the momentum of the truck is exactly equal and opposite the momentum of the car. In this scenario, the work done by the truck is negative in terms of reducing the kinetic energy of the car because the force of the truck and the displacement of the car (centre of mass) are opposite in direction.

In a real situation the momentum of the truck is greater than the opposite momentum of the car and friction is involved. The answer in my last post applies.

[kamil9876]

Evaporade: I think your definition of work is correct(I checked on wikipedia), however you are only considering half the story.

When the car is struck by the truck the work the truck does on the car is negative, however once it turns around the work is positive.
This situation can be compared to a ball being thrown up in the air. Initially grativy does negative work because the displacement is up while the force is acting down, however after the turnaround the work is positive since the direction the ball is falling is down and the force is also down.
In the truck case the car's direction is still positive for som time however beyond that it begins to turn around (just like the ball does) and so the direction of car and force are both negative hence positive work.
This is consistent with /0's claim concerning the change in kinetic energy: car slows down, then speeds up after turn around(just like ball does)

Key: where d is displacement of object work is being done on while F is the force being applied.

Edit: my claim about the car's direction being positive for some small time can be justified as follows: the velocity of the car is +40m/s, however it eventually becomes negative but the change in velocity must be continous because if it wasn't then we would need infinite force.(assuming mass of car is constant: So a change of over 40m/s in v in an instant needs unrealistically large force)

[methodsboy]

Hey does anyone know what the units for are? Are they meant to be or ?

N/m

[evaporade]

I considered the whole story that is why I said 'The answer depends on the mass of the car, mass of the truck, velocity of the truck and whether friction is involved.' The 'overall' work (+ or -) done by the truck on the car depends on its momentum before collision. 'Overall' work was discussed because it is what years 11 and 12 required to know, not moment to moment analysis of the motion.

[evaporade]

'my claim about the car's direction being positive for some small time can be justified as follows: the velocity of the car is +40m/s, however it eventually becomes negative but the change in velocity must be continous because if it wasn't then we would need infinite force.(assuming mass of car is constant: F_{Average}=m\frac{\Delta V}{\Delta t} So a change of over 40m/s in v in an instant needs unrealistically large force)'

I think you are talking nonsense here. For sure the centre of mass of the car slows down gradually due to the collapse of the front end of the car. You don't need 'unrealistically large force' to stop the car.

[evaporade]

I better stop here, otherwise I may end up like the person in /0 's picture. This is to be expected on this forum.

[kamil9876]

I think you are talking nonsense here. For sure the centre of mass of the car slows down gradually due to the collapse of the front end of the car. You don't need 'unrealistically large force' to stop the car.

I didn't say that, of course the car can be stopped, and i said that it is being stopped gradually(my use of the word 'continously') i just said that being stopped in an 'instant' is impossible meaning that it cannot go from 40m/s to some negative velocity without going through all the velocities in between. Now because of that, it is still going forward for some time (and so negative work in this duration).

Btw, i don't mind you saying that what i said was nonsense so I give you the right to continue criticizing any of my nonsense.

[/0]

Oh dear, I think this is getting a bit beyond VCE
So you're saying, if the car starts the collision moving in one direction, but the dollision makes the car move in the other drection, then you have both positive and negative work on the car? How would you deal with that?

[kamil9876]

well its NOT positive AND negative at the same time. Think about the ball being thrown in the air scenario. On the path up its negative, and the path down it's positive. Negative for path up because its a dot product of two vectors opposite to each other. Otherwise for path down.

In the car situation, the moment of time the car is being stopped, it is slowing down and so it's analogous to the path up situation (force is acting opposite to the initial tiny displacement of the car). But once the car begins moving backwards, well that is analogous to the path down situation.

[kamil9876]

and yes, i think this is beyond yr12 since it is formalised with the dot product(and math in yr12 physics is nothing beyond manipulating linear equations). THe 'negative work because kinetic energy decreased' argument get's you the same answer anyway.

I think the answer to this question depends on whether you are considering the work it takes just to stop the car or if the car is being dragged back by the truck(they squashed into one wreck) and you want the work done when the car is being dragged back by the truck.

[/0]

Thanks kamil and evaporade !

For these electrical devices, which must use AC, which must use DC and which can use either:
Resistor
Thermistor
LED
LDR
Photodiode
Phototransistor
Voltage Amplifier

thx

[/0]

I am having lots of problems with the Insight 2009 Trial Exam...
http://vcenotes.com/forum/index.php/topic,8214.15.html

Question 6 - Electronics and Photonics:
If you look at the transfer characteristic of the amplifier, a voltage of should give an output voltage of something like , and a voltage of should give , because that is well within the saturation region of the amplifier.
But in the answer they just apply the linear gain to the max and min of the input graph... How is this correct if it defies the characteristic graph??
I would expect an output graph that is mostly constant at 10V but which drops down every now and then.

Question 11 - Electronics and Photonics:
Are the answers wrong with this!?? Should the modulated signal be modulated at its minimums as well? They just have it at constant minimum!

Question 5 - Structures and Materials:
Isn't D as acceptable an answer as A?