VCE Specialist 3/4 Question Thread!

[Homer]

dont get the answer

[brightsky]

a) (6000-4x)-400g=400a
1500-x-100g=100a
a = (1500 - 100g - x)/100
when x = 20
a = (1500-100g - 20)/100 = 5 m/s^2

b) a = (1500 - 100g - x)/100
d/dx(1/2*v^2) = (1500 - 100g - x)/100
1/2*v^2 = 1/100 int (1500 - 100g - x) dx
= 1/100 (1500x - 100g*x - x^2/2 + c)
we need that v=0 when x=0
c = 0
1/2*v^2 = 1/100 (1500x - 100g*x - x^2/2)
v^2 = 1/50 (1500x - 100g*x - x^2/2)
=x/100 (3000 - 200g - x)
when x = 40
v^2 = 40/100 (3000 - 200g - 40)
= 0.4*(3000 - 1960 - 40)
= 400
so v = +- 20
we know that at x = 40, the crate is still going up
so v = 20
but it doesn't really matter because we want the speed, which is the magnitude of the velocity
so speed = 20 m/s

[availn]

dont get the answer

The net force on the crate is always going to be upwards, and is equal to tension minus weight. Weight is mg = 400·9.8 = 3920N. The formula for tension is given as (6000-4x)N. So your base formula you're working with is:

F_net = 6000 - 4x - 3920 = 2080 - 4x

If we divide everything by 400 (because F=ma), we have an equation for acceleration:

a = 5.2 - 0.01x

So, to find the acceleration at 20m, you just sub 20 into x.
To find the velocity, you'll have to integrate the equation first, then sub in x=40.

[Homer]

thanks guys! forgot about the weight

[barydos]

Apparently the answer is B? It could possibly be wrong I don't know for sure at the moment.

But yeah I can't seem to get to that conclusion.

I thought it'd be something like
(integral from 0 to pi/4 of )

HeLp!

[Phy124]

(Image removed from quote.)

Apparently the answer is B? It could possibly be wrong I don't know for sure at the moment.

But yeah I can't seem to get to that conclusion.

I thought it'd be something like
(integral from 0 to pi/4 of )

HeLp!

I think you were on the right track but chose the wrong line for

edit: I could not knowingly leave what was supposed to be "but" as "bus"

[barydos]

I think you were on the right track bus chose the wrong line for

Oh wow, MY BAD hahah
Thanks a lot man, now my mind is settled enough to go to sleep

[Homer]

what about this one guys?

the answer say it is 12500(9e^(-0.05t) + 1)

i get 137500e^(-0.05t) -12500

[lzxnl]

v=25*(1-e^(-0.05t))
so a=25*0.05e^-0.05t=1.25*e^-0.05t m/s^2
Net force = driving force - resistance = ma
Mass = 100000 kg (this question really needs to specify the units more accurately; units of the resistance force?)
Resistance force = 12500*(1-e^-0.05t) (I am going to assume this is in Newtons)
Driving force = ma+resistance = 10^5*1.25e^-0.05t + 12500 - 12500e^-0.05t
=125000e^-0.05t - 12500e^-0.05t +12500
=12500(10e^-0.05t - e^-0.05t +1)
=12500(9e^-0.05t + 1)

To be honest, the question is pretty bad with its units; that will be very confusing. In general, assume SI units if not specified; forces measured in Newtons, velocities in m/s, acceleration in m/s^2 and masses in kg.

[Homer]

nice thanks

also why i dont understand part b,

dont they act on the same body so not action reaction pair?

[availn]

nice thanks

also why i dont understand part b,

dont they act on the same body so not action reaction pair?

They mean, name the forces that act on the glass, and their reaction pairs.

[Homer]

the answer is A but don't know why and how?

[BubbleWrapMan]

Since it's accelerating slower than , it must be falling and encountering air resistance. If it were travelling up and encountering air resistance, it would be accelerating faster than , since the air resistance would also be forcing it down.

[lzxnl]

It's trippy because air resistance acts against motion. If the ball was moving up, air resistance and gravity would combine to slow the ball down, which would produce an acceleration of more than 9.8 ms^-2 down. As the air resistance appears to weaken the gravitational acceleration, it must be acting against gravity, i.e. object must be moving downwards.

[jono88]

Cross section of a water channel is defined by f(x)=asec(pix/15)+d. The top of the channel is level with the ground and is 10m wide. The channel is 5m deep. Find a and d.