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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: panicatthelunchbar on December 28, 2011, 10:34:52 pm

Title: Help!
Post by: panicatthelunchbar on December 28, 2011, 10:34:52 pm: Hey guys, my physics is very weak, can someone please explain how they would solve this problem!?

A cyclist rolls freely from rest down a slope inclined at 20 degrees to the horizontal. The total mass of the bicycle and cyclist is 100kg. The bicycle rolss for 12 seconds before reaching the horizontal surface. The surface exerts a constant friction force of 300N on the bicycle.

a) What is the net force on the bicycle (including the cyclist) [42N]
b) What is the acceleration of the bicycle? [0.42 m/s2]
c) what is the speed of the bicycle when it reaches the horizontal surface? [5 m/s]

Thanks
Title: Re: Help!
Post by: samad on December 28, 2011, 10:55:44 pm: a) as soon as you read "net force", you should consider the vector addition of forces. since the cyclist is moving down the incline, there is no motion perpendicular to the incline, hence the net force in the perpendicular direction is zero. the net force in the parallel-to-incline direction is the vector sum of the parallel component of the cyclist's weight (Wp) and the friction (F). The weight of the cyclist is the force that the earth exerts on him due to gravity, W=mg=100g where g is the acceleration due to gravity (10 ms-2). to find Wp, we need to use trigonometry (nearly every inclined plane question needs this).
now, Wp= 100gsin(20 deg). drawing a labelled force diagram and figuring out the angles will help u understand where this comes from.
hence Fnet= Wp-F
=100*10*sin(20 deg)-300
= 42 N

b) Here, our answer to part a helps. recall that Fnet=ma where m is the mass of the object and a is the acceleration. we know m and Fnet, so we can transpose for a: a=Fnet/m.
hence a=42/100
=0.42 ms-2

c) whenever a distance or time taken to move along an incline comes up, think "constant acceleration formulae" this is what we need to use here. we know:
u= 0 ms-1
t= 15 s
a= 0.42 ms-2
v= ?

to figure out which one of the five to use, always look at what we don't need at all. here, we neither know nor need to know the distance travelled (s or x), hence use the formula without s:
v=u+at
sub in:
v= 0+0.42*12
= 5 m/s

conceptually, the cyclist starts from rest and gains 0.42 metres per second in speed every second. hence after 12 seconds its speed will be 0.42*12.
Title: Re: Help!
Post by: panicatthelunchbar on December 29, 2011, 01:18:35 pm: That is a great explanation, thanku so much! :D
Title: Re: Help!
Post by: nina_rox on December 29, 2011, 07:25:55 pm: Could someone explain question three on the 2005 VCAA exam (http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/05physics1.pdf)

I don't understand why you subtract m x g from the retarding force and how they get the mass?
Title: Re: Help!
Post by: abd123 on December 29, 2011, 07:40:48 pm: Quote from: nina_rox on December 29, 2011, 07:25:55 pm
Could someone explain question three on the 2005 VCAA exam (http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/05physics1.pdf)

I don't understand why you subtract m x g from the retarding force and how they get the mass?
I guess the equation for spring transformed into gravitational potential energy? I'm not to sure about it.

1/2kx^2=mgh, i feel like i might be wrong.

Hoped i helped.
Title: Re: Help!
Post by: Lasercookie on December 29, 2011, 08:16:47 pm: You're correct abd123.

You can say that energy is conserved, so therefore:

Kinetic Energy = Elastic Potential Energy = Gravitational Potential Energy.

You know from question 1 that the Elastic Potential Energy = 100J. You also know the mass - that is given to you in the original paragraph (0.20kg). You also know g = 10.

$100 J = mgh$

Sub in m and g, solve for h.

The other way I can think of solving this is to use the equations of constant acceleration (as you have initial velocity from question 2).

$v^2 = u^2 + 2ax \rightarrow x = \frac{v^2 - u^2}{2a}$
$x = \frac{v^2 - u^2}{2a}$

Sub in $v = 0, u = \sqrt{1000}, a = 10$ and you have your answer.

Both methods give you the answer of 50.

edit: fixed LaTeX error
Title: Re: Help!
Post by: Phy124 on December 29, 2011, 08:20:16 pm: Quote from: nina_rox on December 29, 2011, 07:25:55 pm
Could someone explain question three on the 2005 VCAA exam (http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/05physics1.pdf)

I don't understand why you subtract m x g from the retarding force and how they get the mass?
You could do v²=u²+2as

Where;
v = 0 m/s
u = answer to Q 2
a = gravity (10 m/s²)
s = the answer ;)

edit: laserdd edited into his post whilst I wrote this, awkward...
Title: Re: Help!
Post by: nina_rox on December 29, 2011, 08:40:58 pm: Ohhh! Thank you guys so much! I realized where things went wrong. I was doing 2012 checkpoints and they didn't give the mass of 0.2kg in the opening paragraph like in the exam. The authors must of accidentally skipped that important information! Thanks all for your help!