Dekoyl's Questions

[dekoyl]

I just encountered a centre of mass question which I'm unfamiliar with and has never been mentioned by my teacher.
Are these type of questions in the course? The question also requires the equation:



Thank you

[TrueTears]

Hmmm I met one of these questions in the Jacaranda book, and only ONE lol

I don't think centre of mass is in the course, but it's good to know, not hard to remember the equation.

[/0]

Haven't read the 2009 study design but I've seen questions on them in 2009 Jacaranda and also I think you're quite likely to use it in Structures and Materials.

[dekoyl]

Hmmm I met one of these questions in the Jacaranda book, and only ONE lol

I don't think centre of mass is in the course, but it's good to know, not hard to remember the equation.

Yeah I'm referring to that ONE. =P

Haven't read the 2009 study design but I've seen questions on them in 2009 Jacaranda and also I think you're quite likely to use it in Structures and Materials.

Ah okay we have a different DS =]

Thanks TT and /0.

[dekoyl]

Hmm.

We are given a scenario with a car towing a trailer (with a pull bar).
The car has a constant driving force of 2400N. The car experiences 400N friction and the trailer experiences 200N friction.

We're told to calculate the work done by the car over a distance.
Using , for F I used 2400 - 400 - 200 = 1800N. However in the solutions, they used F = 2400

Am I wrong? I thought it was . Was ?

Thanks

[TrueTears]

Yes, answers are right, you gotta becareful here because it says the work DONE BY the CAR. It does not matter whether there's friction or anything else acting on the car, but only the forces that the CAR ITSELF exerts.

[dekoyl]

Yes, answers are right, you gotta becareful here because it says the work DONE BY the CAR. It does not matter whether there's friction or anything else acting on the car, but only the forces that the CAR ITSELF exerts.

Argh the language.

Thanks man. TrueTears to the rescue again

[dekoyl]

In physics, if you don't use the method assessors expect, would you lose marks even if you get the solution right?

I encountered a question where the knowledge of needed to be applied to calculate velocity but I used and got the correct answer. As long as the correct answer is obtained, any method is fine right?

Thanks

[TrueTears]

In physics, if you don't use the method assessors expect, would you lose marks even if you get the solution right?

I encountered a question where the knowledge of needed to be applied to calculate velocity but I used and got the correct answer. As long as the correct answer is obtained, any method is fine right?

Thanks

yeap as long as your answer is correct, give your working method makes sense and is also logical, then you still get the marks.

If you get the answer but your working was not logically, then you won't get the method marks.

[dekoyl]

Alrighty =] Thanks again TT.

I'm asking all these questions because I have a SAC on Monday. Yep =|

[Over9000]

Impulse = , so you can use both (whichever one you have more values for)

[TonyHem]

I thought its the same as the formulas u mentioned above, it just depends on the information given/known.
I think..

[dekoyl]

First one I'm not sure what I've done wrong.
A 20kg wagon is pulled along the level ground by a rope inclined at above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with an acceleration of ?

Okay - after taking the friction and angle of the rope into account, I end up with:


The answer, however, is .


Second one: I'm just not sure what to do as I can't picture the question =S
An inclined plane making an angle of with the horizontal has a pulley at its top. A 30kg block on the plane is connected to a freely hanging 20kg block by means of a cord passing over the pulley. Compute the distance the 20kg block will fall in 2s starting from rest. Neglect friction.

Thanks everyone.

[Over9000]

For q.1
F=ma
F=
F= 8N
so the total net force is 8N
if friction force is 30N, there must be a force of 38N acting in the opposite direction (i.e direction of pulling) horizontally, however we must find the pulling force at to horizontal
Therefore the force is (draw a triangle if you dont understand this part)
This gives 43.8786N (4 d.p)
Which is 43.9N (1 d.p)

[dekoyl]

^Ah crap. My answer to the first question was a fluke then

Thanks Over9000 Didn't expect anyone to help at this hour