Amatuer Physics Questions

[stonecold]

I am trying to self learn physics and it is not going very well.  This is so sad.  I did year 11 physics and went quite well in it.  Less then two years later and I cannot remember squat haha.

I have no idea how to solve this question.  I get up to having force = 1250 N, then I have no idea how to do the conversion to work out the grams.

Help please??

[Thu Thu Train]

F = ma
m = F/a

F is force in Newtons(N)
m is mass in kg
a is acceleration in m/s²(meters per second per second)


Really need more info though.

[stonecold]

This is what I have so far...

v²=u² + 2ax

v=500
u=0
x=0.5
a=?

500²=0² + 2(0.5)a

a=250000

F=ma

m= 5g = 0.005 kg

F=0.005 x 250000 = 1250 N

Then, I am screwed.  :/

[stonecold]

shit, sorry...i forgot to post the questions

i am such an idiot...doing it now...

edit:  okay done, now people might understand what I am talking about :p

[Thu Thu Train]

It's been a long time since I've done kinematics so someone else can check my work for me but:


Your force of 1250N is correct and the question tells us that there is 100N/cm²/g of force applied to the bullet.

The surface area of the bullet is 2.545cm² and so we can determine that for every gram of gunpowder there is 254.5N of force applied to the bullet (100N/cm²/g * 2.545cm²)

and now the easiest part comes:
We know that the bullet has a force of 1250N applied to it whilst leaving the gun.


which rounds to 5g

Q.E.D

[stonecold]

It's been a long time since I've done kinematics so someone else can check my work for me but:


Your force of 1250N is correct and the question tells us that there is 100N/cm²/g of force applied to the bullet.

The surface area of the bullet is 2.545cm² and so we can determine that for every gram of gunpowder there is 254.5N of force applied to the bullet (100N/cm²/g * 2.545cm²)

and now the easiest part comes:
We know that the bullet has a force of 1250N applied to it whilst leaving the gun.


which rounds to 5g

Q.E.D

I'm glad you get it. 

How did you get the surface area?  I am probably missing something ridiculously simple, but I could not work out how to get it.

[Thu Thu Train]

Well you assume the bullet is a sphere and they give you the diameter of the sphere (9mm = radius 4.5mm) then you use the equation to calculate the surface area of the sphere

[stonecold]

Well you assume the bullet is a sphere and they give you the diameter of the sphere (9mm = radius 4.5mm) then you use the equation to calculate the surface area of the sphere

Thanks.  For a split second I thought maybe you assume it is a sphere, but I think it is kinda unreasonable that they don't mention the shape. 

When I think "bullet", this springs to mind:



:p

Anyway, appreciate the help. 

[stonecold]

Can someone please explain why weight force and normal force are not considered an action reaction pair, even when they are directly opposite one another?  (is it something to do with the reaction pair for gravity being between an object and earth?)

Also, what is the difference between gravitational mass and inertial mass?  (even though I have been told they are always the same)

Thanks.

[tony3272]

This is just for the first question. (I've never heard of gravitational/inertial mass)
But the reaction force of the force of gravity that the earth exerts on an object is simply the force of gravity that the object exerts on the earth.
The reason this is not the same as the weight force is best done through an example. If you have a block on a table that has a mass of 10kg, its weight force is 100N. If you then place another block of mass 10kg on top, the weight force of the first block is still 100N, however the normal force exerted will be 200N due to both blocks.
The weight force of an object is the force gravity imposes on it, whereas the normal force is not. The normal force is just the perpendicular reaction force of the object pressing against a surface.

[moekamo]

weight force is the force(gravitational) of the earth on an object and the normal force is also a force by the earth(surface) on an object. Action reaction pairs must point in opposite directions, that is true, but they must also act on DIFFERENT bodies, therefore weight force and normal force are not an action reaction pair, they both act on the object...


Gravitational mass is , which is a re-arranged version of newtons law of gravitation. In other words, it is the mass that exerts/experiences a certain force due to the attraction between another mass() at a distanct .

Inertial mass is , which is essentially the how hard something is to accelerate when a force is applied, larger mass means harder to accelerate, i.e. lower acceleration...

It turns out that both m's have the same value to some crazy accuracy value that is really the precision of the instruments used, but apparently there is no reason that they should have the same value, it just turns out that they do(according to my physics lecturer).

[stonecold]

Thanks, I think I get it now.  So basically, weight force and normal force are never an action/reaction pair because they are both exerted on the same object?  (Whereas in an action/reaction pair, object 1 exerts a force on object 2, and the object 2 exerts an equal and opposite force on object 1)

And I am not even sure that you cover inertial/gravitational mass in VCE :S

Edit:  Cheers guys.  100% makes sense now. 

I will probably we back with more dumbass questions seeing as we have an assessment this Wednesday.

[pi]

And I am not even sure that you cover inertial/gravitational mass in VCE :S

We use that formula, but not in the scenario that you have to

[stonecold]

Struggling with parts c) and d) 

   ...Wednesday is not looking good.   The answers are:

c) is 798 N
d) is 588 N

All I understand is that the weight force is 588 N down, and that at constant velocity the sum of the forces is 0 N.

I worked out that to be accelerating 2.5 m/s/s up, you need a net force of 150 N up, but then I am stumped.  :/

[mardat]

For part c, the two forces are gravity acting downwards from the centre of the person (weight is not given at this point in the question, so I'd assume you simply label the forces with the type of force) and the normal force of the lift upon the person acting upwards from the floor of the lift. The net force on the person will be upwards, as the person is accelerating upwards.

For part d, the force the person exerts on the floor will be the same as the force the floor exerts on the person, i.e. the force required to resist gravity and to accelerate the person at 2.5ms^-2. Therefore F=mg+ma=m(g+2.5)=60*12.3=738N
After the first two seconds, when velocity is constant, the force the person exerts on the floor is just that of gravity, as there is no acceleration. F=mg=60*9.8=588N