Simple Physics

SaBBz · 18 Apr 2006 at 14:40

anksta said:
the 2 components, i.e. the acceleration of the car and the weight of the car acting downwards are at 90 degrees to eachother so there would be no resultant anywya.

Also, there is no force in the vertical direction because the car isnt moving up or down, therefore, the weight of the car is being equally opposed by the road, you only have to worry about the horizontal direction, which has already been explained is simply F= ma.

I'm glad someone pointed this out. You cannot just ignore the weight of the car if the car is going down/up a slope

(The OP might have to do a question like this at some point

)

benjo plz. · 18 Apr 2006 at 14:41

Arcade Fire said:
No, the work done is the change in kinetic energy.

In this case, afaik the change in KE = the resultant force and presumably the work done...

daz · 18 Apr 2006 at 14:42

SaBBz said:
I'm glad someone pointed this out. You cannot just ignore the weight of the car if the car is going down/up a slope (The OP might have to do a question like this at some point )

That's how Newton approximated G IIRC... by rolling objects down inclines.

Tommy B · 18 Apr 2006 at 14:43

God hates me. This is SO easy yet I can't do it. I'm actually going crazy now!!

SaBBz · 18 Apr 2006 at 14:43

Phnom_Penh said:
In this case, afaik the change in KE = the resultant force and presumably the work done...

They aren't the same thing. Force /= energy

Visage · 18 Apr 2006 at 14:47

Phnom_Penh said:
Shouldn't the resultant force = change in ke...

No. 2 seconds thought shows that this cant be true, as the left hand side is in Newtons and the right hand side is in Joules.....

Tommy B · 18 Apr 2006 at 14:49

I'll send a quid by paypal to the first person who tells me how to do this **** :mad:

Arcade Fire · 18 Apr 2006 at 14:51

Tommy B said:
Seriously I'm now stuck on part two.

Mass of car = 1.3x10^3 kg

b) When the accelerating car reaches a speed of 2.2ms-1, the total force opposing the motion of the car is 410N. Calculate the driving force provided by the wheels.

I don't think that you can solve this without more information. For example, is the car still accelerating at a constant rate of 2m/s^2? Even then the question is made much trickier by the fact that you're now considering an opposing force, and the way that the question is worded makes it sound as though it's a variable force.

Can you type out the whole question? Visage, do you agree with me?

benjo plz. · 18 Apr 2006 at 14:51

Visage said:
No. 2 seconds thought shows that this cant be true, as the left hand side is in Newtons and the right hand side is in Joules.....

yeh, fair point. You were beaten to it though.

Hey TommyB, this isn't from an AQA textbook is it?

SaBBz · 18 Apr 2006 at 14:53

Phnom_Penh said:
yeh, fair point. You were beaten to it though.

Hey TommyB, this isn't from an AQA textbook is it?

If you want to work out work done you find the integral of the force with respect to the distance moved iirc. (WD = F.dr (where the dot is a dot product))

Tommy B · 18 Apr 2006 at 14:53

Arcade Fire said:
I don't think that you can solve this without more information. For example, is the car still accelerating at a constant rate of 2m/s^2? Even then the question is made much trickier by the fact that you're now considering an opposing force, and the way that the question is worded makes it sound as though it's a variable force.

Can you type out the whole question? Visage, do you agree with me?

With pleasure...

5) A car accelerates at a steady rate of 2.5ms-2 along a straight, level road. The mass of the car is 1.3X10^3 kg.

a) calculate the magnitude of the resultant force acting on the car?
F = MA = 1.3X10^3 * 2.5 = 3250N

b) When the accelerating car reaches a speed of 2.2ms-1, the total force opposing the motion of the car is 410N.

Calculate

1) The driving force provided by the wheels.
2) The power delivered to the wheels of the car.

benjo plz. · 18 Apr 2006 at 14:54

SaBBz said:
If you want to work out work done you find the integral of the force with respect to the distance moved iirc.

yep.

daz · 18 Apr 2006 at 14:55

Arcade Fire said:
I don't think that you can solve this without more information. For example, is the car still accelerating at a constant rate of 2m/s^2? Even then the question is made much trickier by the fact that you're now considering an opposing force, and the way that the question is worded makes it sound as though it's a variable force.

Can you type out the whole question? Visage, do you agree with me?

It depends if the force opposing the motion is a function of v or v^2 or constant, and whether the car is intending to carry on accelerating.

SaBBz · 18 Apr 2006 at 14:57

daz said:
That's how Newton approximated G IIRC... by rolling objects down inclines.

I'm doing a physics degree at Nottingham uni and I never knew that

Thanks

Tommy B · 18 Apr 2006 at 14:58

SaBBz said:
I'm doing a physics degree at Nottingham uni and I never knew that Thanks

Really? Do you mind helping me then. I have offered a whole one pound.

benjo plz. · 18 Apr 2006 at 14:58

Tommy B · 18 Apr 2006 at 15:01

C'mon that's 95p after fees. Think of what you could buy with that!

Arcade Fire · 18 Apr 2006 at 15:01

Tommy B said:
With pleasure...

5) A car accelerates at a steady rate of 2.5ms-2 along a straight, level road. The mass of the car is 1.3X10^3 kg.

a) calculate the magnitude of the resultant force acting on the car?
F = MA = 1.3X10^3 * 2.5 = 3250N

b) When the accelerating car reaches a speed of 2.2ms-1, the total force opposing the motion of the car is 410N.

Calculate

1) The driving force provided by the wheels.
2) The power delivered to the wheels of the car.

Ah okay, that makes it much nicer. All you need to know for part (b) is that the resultant force is 3250 N, and that the force opposing the motion is 410 N. So if the driving force is D, then you just need to do

D - 410 = 3250
D = 3660

and for the next part, you use P=Fv, or Power = Force x Velocity, with a force of 3660 N and a velocity of 2.2 m/s.

SaBBz · 18 Apr 2006 at 15:03

F - Drag = 3250N
Drag = 410N
Therefore F = 3660N

P = F*v (v = 2.2m/s)

P = 8052W?

[edit] Beaten to it

[/edit]

Tommy B · 18 Apr 2006 at 15:04

Arcade Fire said:
Ah okay, that makes it much nicer. All you need to know for part (b) is that the resultant force is 3250 N, and that the force opposing the motion is 410 N. So if the driving force is D, then you just need to do

D - 410 = 3250
D = 3660

and for the next part, you use P=Fv, or Power = Force x Velocity, with a force of 3660 N and a velocity of 2.2 m/s.

Cheers. If you want the quid email me your paypal addy