Author Topic: NEAP 09?!!!!!! (Read 2158 times) Tweet Share

Stroodle · « **Reply #15 on:** June 05, 2010, 07:17:48 pm »

If you just used the force of the guy pushing you wouldn't be taking the friction of the grass into account.
It would be correct if there was no opposing forces though; cause then there would be no other horizontal force besides the guy pushing.

Keyzer · « **Reply #16 on:** June 08, 2010, 10:00:29 am »

Quote from: m@tty on June 05, 2010, 12:29:04 pm

$t=4.59s$ $u=0$ $s=10m$ v=? a=?

$s=ut+\frac{1}{2}at^2$

You need to do it this way to get the net force otherwise you are calculating the work done by the man pushing.

$a=\frac{2s}{t^2}=\frac{20}{(4.59)^2}=0.9493 ms^{-2}$

$F=ma = 0.9493\times 30=28.48 N$

$W=Fd=28.48\times 10=284.8 J$

EDIT: 1900

I see what you did there, but it says accelerates from rest to 10 metres? So shouldn't that mean that v = 10 m/s? Or do they mean that the distance is 10 m. Damn ambiguity.

m@tty · « **Reply #17 on:** June 08, 2010, 10:20:26 am »

Nah, it says 10 m. The unit is there. Worded a little wonky though.

FatnessFirst · « **Reply #18 on:** July 26, 2010, 09:41:35 pm »

http://vcenotes.com/forum/index.php/topic,19716.0.html

Blakhitman · « **Reply #19 on:** July 26, 2010, 09:57:35 pm »

Quote from: FatnessFirst on July 26, 2010, 09:41:35 pm

http://vcenotes.com/forum/index.php/topic,19716.0.html

Thanks, but this thread was for unit 3 and that exam is for unit 4

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Author Topic: NEAP 09?!!!!!! (Read 2158 times) Tweet Share

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