VCE Physics Question Thread!

[Shadowxo]

But why do you multiply 7*10^6 by 240 kg? Isnt the graph a force vs distance graph not a gravitational field strength vs  distance graph

It says on the graph that it's the force acting on 1kg, aka the gravitational field strength, which is why you have to multiply the weight.

[Gogo14]

It says on the graph that it's the force acting on 1kg, aka the gravitational field strength, which is why you have to multiply the weight.

*repeatedly slams head on desk*

[Shadowxo]

*repeatedly slams head on desk*

Haha dw, it's the little things that get you (and me)

[Gogo14]

For question 1, the answer is A, but isnt the travelator doing work?

For question 9, how do you do it?

[Gogo14]

Couldnt attach the other photo, but
How do you question 8, really lost
Thanks heaps!

http://imgur.com/uZccyZh

[Shadowxo]

For question 1, the answer is A, but isnt the travelator doing work?

For question 9, how do you do it?

1. For the travelator, there is no acceleration as everything is going at the same speed, therefore F = 0 therefore work = 0
9. You can use Work = Fd
F = ma, a = 9.8, F = .8*9.8 = 7.84
W = Fd = 7.84*1.9 = 14.90

Alternative Way:
Net work done = change in kinetic energy
You should resolve the speed into vertical and horizontal components.
v = 30 ms-1
Horizontal speed = 30cosx = 21.21 ms-1
Vertical speed = 30sinx = 21.21 ms-1
Horizontal speed is unchanging (no force acting on it)
s = -1.9 u = 21.21 a = -9.8 v = ?
v² = u² + 2as
v² = 450 + 37.24
v = 22.07 ms-1 down
net v = 30.61
∆ke = final ke - initial ke
 = 374.90 - 360 = 14.90

[Gogo14]

1. For the travelator, there is no acceleration as everything is going at the same speed, therefore F = 0 therefore work = 0
9. You can use Work = Fd
F = ma, a = 9.8, F = .8*9.8 = 7.84
W = Fd = 7.84*1.9 = 14.90

Alternative Way:
Net work done = change in kinetic energy
You should resolve the speed into vertical and horizontal components.
v = 30 ms-1
Horizontal speed = 30cosx = 21.21 ms-1
Vertical speed = 30sinx = 21.21 ms-1
Horizontal speed is unchanging (no force acting on it)
s = -1.9 u = 21.21 a = -9.8 v = ?
v² = u² + 2as
v² = 450 + 37.24
v = 22.07 ms-1 down
net v = 30.61
∆ke = final ke - initial ke
 = 374.90 - 360 = 14.90

Thanks, for question 9, the answer says its 374J ?

[lzxnl]

Thanks, for question 9, the answer says its 374J ?

I get the same answer as Shadowxo.

[Shadowxo]

Thanks, for question 9, the answer says its 374J ?

I think they were treating the d in W=Fd as distance not displacement (although I believe displacement, not distance, should be used)

[lzxnl]

To properly resolve this issue, let's go back to the VERY basics. Which are probably not covered in much depth in VCE, but whatever.

What is work? Work is the ability of an object to exert a force over a displacement parallel to the force. Any object that is capable of doing work then has energy (this is my preferred definition of energy). Why is this directional requirement important? Defining it this way allows energy to be mathematically conserved, which makes it a physically useful definition.

So, if work is exerting a force in the direction of motion, forces that are constantly perpendicular to an object's motion should not do work. And indeed, uniform circular motion arises when the net force is ALWAYS perpendicular to the object's motion (and satisfies a magnitude requirement); the speed and thus the energy do not change in such a motion.

Now, because 2D motion can be separated into its two independent dimensions, we can consider the net force, the gravitational force, as acting on the horizontal and vertical components separately. The gravitational force is perpendicular to the horizontal velocity component, so no work is done there. It only affects the vertical velocity component. Therefore, as far as gravity is concerned, the object is only moving vertically. (it is possible to find a frame of reference such that the object is indeed only moving vertically; this frame of reference is best pictured by someone running straight under the object), and so the work it does on the object should only involve the vertical displacement.

Final question. Do we use displacements or distances? Well, suppose you throw a ball up. Gravity is clearly slowing it down when the motion is opposing the direction of gravity, and speeds it up when the motion is in the direction of gravity. Therefore, the direction of motion is important and you use displacements. It is possible to prove (I won't though) that gravity is a conservative force, which in this problem means that you only need to know the initial and final state to calculate the work; the path doesn't matter. In other words, you only need to know that the ball had a vertical displacement of 1.9 m. Any calculation involving the distance displays a fundamental misunderstanding of physics. Indeed, calculating the actual distance through the air travelled by the javelin along its parabolic trajectory isn't trivial; I invite any spesh students to do so as a preview of the coming calculus topics.

[-273.15]

Helloo,
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept

Thank you!

[Syndicate]

Helloo,
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept

Thank you!

This pdf is quite helpful.

[lzxnl]

Helloo,
Could someone please explain to me how the relationship between normal force and weight force changes when in an elevator (i.e. accelerating up and down) and could you also explain WHY because I cant really understand this concept

Thank you!

Hi there,

Suppose I want to lift a box. Suppose I want to lift it up fast. Then, the vertically upwards net force needs to be large. This means that the force I'm applying to the box has to be a lot larger than its weight.

Suppose I want to lower a box. Suppose I want it to fall fast. Then, the vertically downwards net force needs to be large. This means that the force I'm applying to the box has to be a lot smaller than its weight.

Now replace lifting/lowering the box with the lift applying a normal reaction force onto the box, which plays the same role. Does that help? The normal force is functionally equivalent to the force applied by a person onto said box.

[-273.15]

Hi there,

Suppose I want to lift a box. Suppose I want to lift it up fast. Then, the vertically upwards net force needs to be large. This means that the force I'm applying to the box has to be a lot larger than its weight.

Suppose I want to lower a box. Suppose I want it to fall fast. Then, the vertically downwards net force needs to be large. This means that the force I'm applying to the box has to be a lot smaller than its weight.

Now replace lifting/lowering the box with the lift applying a normal reaction force onto the box, which plays the same role. Does that help? The normal force is functionally equivalent to the force applied by a person onto said box.

Thank you so much !

[Guideme]

1. Explain the importance of keeping a lid on a simmering saucepan of water in terms of latent heat of vapoiurisation?

2. Explain in terms of the kinetic particle model why you can put your hand safely in a 300C oven for a few seconds, while if you touch a metal tray in the same oven your hand will be burned

3. How does evapouration of water cause a reduction i the temperature of the surrounding air?

I am findinng Physics very difficult and my teacher just reads from the textbook, so it would be nice if anyone can provide an answer and explain it in detail so i can understand it thank you!