Bazza's 3/4 Physics question thread

[WhoTookMyUsername]

What does "average net force" refer to?
I can't find the specific formula in my textbook but i'm pretty sure in general it just means
is this a real physics formula or just a common sense thing?
Is the unit or just N (btw textbook says "N", i thought Ns-1 was more logical)? and why?

thanks

[abd123]

Average net force, i guess  the added vectors ?

the netforce over time applied, it might have came out of the momentum/impulse formulae ,
i think it implies the impact of the body.

its either way.

Yes the unit is , its the unit for

I might be wrong lol.

Hoped I helped

[samad]

the unit is N- your definition is slightly incorrect here.

"Average net force" appears in situations where there is a time-varying force applied to an object, which causes a net change in that object's momentum (i.e the force has a net impulse)

You could calculate the net change in momentum using


If you knew the time over which the varying force was applied, you could then calculate F. this would return the average net force, with unit N.

You can consider this with knowledge of maths methods. the graph of the applied force against time is a function. the area under that function is the impulse. the average net force is the average value of the function (the height of a rectangle that would give that same area).

[moekamo]

im pretty sure the average net force is just the net force, for example, i may apply a force to a block to accelerate it across a room, i'd start off with a large force then it would decrease as the block started moving, but on average over the entire distance or time there would have been an average net force of say 10N or whatever.

It's just to make problems simpler so you don't have to worry about changing forces, by giving you the average you can just use F=ma as usual.


I could be wrong, but this is my interpretation...

EDIT: beaten, and samad's answer seems more rigourous, but im pretty sure we're essentially saying the same thing

[WhoTookMyUsername]

thanks for the help

another question (a lot of my questions for all subjects will be simple, completely neglected 1/2s this year :-/)

a linear graph with distance as x value and kinetic energy as y value is shown with the points (2,5) (5,2) (gradient = -1) shown

What is the magnitude of the frictional force that acts on the puck when it is on the rough surface?

I'm not sure what to do here... well, i know it's something to do with a work formula, i did W = fxcos(a) (where cos (0) = 1) which got me the correct answer, but i really have no idea why it worked, could someone explain why this formula relates to this question? what is frictional force? force applied per second? thanks

[xZero]

w=f*x, let me transform this into this equation so you understand it better, w=f*x => y=m*x. y is your work or kinetic energy, m is your gradient, in this case its f(force) and x is the distance travelled. So by finding the gradient you're essentially finding the frictional force

[paulsterio]

Magnitude of the frictional force?
Well firstly, W = F.x = |F| |x| cos(a)

Now just substitute two points on the curve in for W and x and you should be able to solve for F. Since the force is in the same direction as the motion, the angle a = 0.

[nisha]

I'm not sure if I can ask this question here, but i'll give it a shot:)
Basic Motion question, but I forgot completely how to do it.

"A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80km/hr in 10.0 seconds and reaching a constant speed of 100km/hr after a further 5.0 seconds. At what time will the policeman catch up with the car?

Thanks!

[moekamo]

draw a velocity-time graph for both car and motorcycle on the same graph, then equate the area underneath to be the same when time=t, i.e. express the distance each vehicle travels in terms of time from when the car overtakes the motorcycle.

then solve for t

Also, make sure you convert all the speeds to m/s

[Phy124]

I'm not sure if I can ask this question here, but i'll give it a shot:)
Basic Motion question, but I forgot completely how to do it.

"A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80km/hr in 10.0 seconds and reaching a constant speed of 100km/hr after a further 5.0 seconds. At what time will the policeman catch up with the car?

Thanks!

I knew someone else would ask!

I've written you a fairly detailed answer because many students seem to struggle with this question.

So once you've had a go, if you still can't work it out have a look at this.

Or beforehand if you wish, no one's stopping you...

(I'm quite certain it's the write answer - rings a bell)

p.s. in future it's probably best to make your own thread to ask questions in rather than using Bazza's

[nisha]

Thankyou! Yes that I was the right answer:)

I will keep that in mind

[WhoTookMyUsername]

A rubber ball of mass 80g bounces vertically on a concrete floor. The ball strikes the floor at 10ms-1 and rebounds at 8.0 ms-1

7a) calculate the average net force acting on the ball during its contact with the floor
b) calculate the average force that the floor exerts on the ball

now a) is 29 N up and b) is 30 N up...

I'm confused, i thought they should be the same?

An explanation in english (conceptual) using this question as a reference?

thanks

[moekamo]

average net force includes the gravitational force acting downwards, which 'cancels' with some of the force upwards from the floor, hence average net force is slightly smaller than the force exerted by the floor...

[Lasercookie]

7a is asking about about the net force. The net force is the vector sum of all the force in the system blahblahblah.
7b is asking specifically about the force the floor exerts. This force that is exerted by the floor alone isn't the net force. This is why the answers for 7a and 7b are different.

7a.
Sub in etc. you know how to do the calculations.
You end up getting 28.8N which rounds up to 29N.

7b. There are two forces in this system - these comprise the net force:
The force of gravity on the ball is (note that it acts downwards)

N is the normal reaction force, it acts upwards. This is the force of that the floor exerts on the ball.

which rounds up to 30N

[Aurelian]

thanks laseredd

With the formula F=-kx

why is the negative there? is it assuming the displacement is negative :S?

The negative is just to indicate that the force exerted by the spring is opposite in direction to the force exerted on the spring.