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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Glockmeister on August 18, 2008, 07:41:20 pm

Title: A question
Post by: Glockmeister on August 18, 2008, 07:41:20 pm: A particle slides from rest down a rough plane inclined at 60 degrees to the horizontal. Given that the coefficient of friction between the particle and the plane is 0.8, find the speed of the particle after it has travelled 5m.
Title: Re: A question
Post by: phagist_ on August 18, 2008, 07:43:23 pm: Have you drawn a free body diagram with all the forces acting on it?

Then it is only a matter of getting the acceleration from the net force then applying constant acceleration formulae.
Title: Re: A question
Post by: Glockmeister on August 18, 2008, 07:57:44 pm: Done that, didnt get the answer they have in the book.
Title: Re: A question
Post by: Mao on August 18, 2008, 08:02:50 pm: forces:
gravity: component in the direction of the slope $F_d=mg\sin \theta$
component perpendicular to the slope (equal to the normal reaction force) $F_N=mg \cos \theta$
friction: in the direction opposing the velocity (i.e. upwards parallel to the slope) $F_f=\mu F_N=\mu mg \cos \theta$

acceleration due to gravity: $a_d=\frac{F_d}{m}=g\sin 60^o=\frac{\sqrt{3}g}{2}$

deceleration due to friction: $a_f=\frac{F_f}{m}=\mu g\cos 60^o=0.4g$

net acceleration: $a_{net}=a_d-a_f=\left(\frac{\sqrt{3}}{2}-\frac{2}{5}\right)g=\frac{5\sqrt{3}-4}{10}g$

after 5 meters:

$v^2-u^2=2ad$

$\implies v^2=(2)\cdot \frac{5\sqrt{3}-4}{10}g \cdot (5)$

$\implies v=\sqrt{(5\sqrt{3}-4)g}\approx 6.76ms^{-1}$ (i hope?)
Title: Re: A question
Post by: Glockmeister on August 18, 2008, 08:21:14 pm: Cheers Mao. I somehow left out the g out of the cos... somehow
Title: Re: A question
Post by: /0 on August 18, 2008, 10:02:20 pm: Alternatively with physics we might have

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

In this case $g = 9.8\sin{60^{\circ}}-(0.8)(9.8)\cos{60^{\circ}}}$ and $h=5$ because we only need the 'potential' component down the slope:

$\therefore v = 6.76ms^{-1}$
Title: Re: A question
Post by: Mao on August 18, 2008, 10:13:26 pm: ahhh DivideBy0, you should realise they are exactly the same :P

$v=\sqrt{2gh}\equiv v^2=2ad$