Force on inclined slope (1 Viewer)

jaccs · Aug 30, 2021

Hi, i was given this question, and i am not sure it is even in the syllabus, as i couldn't find anything in a textbook as an example.
i have managed to do part i and ii, but for part three, I ended up with
$\frac{4}{5}\sqrt {10 + 5\sqrt 3 }$
instead of the required
$\frac{4}{5}\sqrt {20 + 10\sqrt 3 }$

Any help would be appreciated

notme123 · Aug 30, 2021

$ma=T-mg\sin60$
$a=10-5\sqrt{3}$
$v=(10-5\sqrt{3})t$
$x=\frac{1}{2}(10-5\sqrt{3})t^2$
$x=16:$
$16=\frac{1}{2}(10-5\sqrt{3})t^2$
$32=(10-5\sqrt{3})t^2$
$\frac{32}{(10-5\sqrt{3})}=t^2$
$\frac{32(10+5\sqrt{3})}{25} = t^2$ (rationalise)
$t^2 = \frac{16(20+10\sqrt{3})}{25}$
$t=\sqrt{\frac{16(20+10\sqrt{3})}{25}}$
$\therefore t=\frac{4}{5}\sqrt{20+10\sqrt{3}}$
BTW this is more of a physicsy questions, if it was truly 3u there would be more vector notation.

idkkdi · Aug 30, 2021

notme123 said:
$ma=T-mg\sin60$
$a=10-5\sqrt{3}$
$v=(10-5\sqrt{3})t$
$x=\frac{1}{2}(10-5\sqrt{3})t^2$
$x=16:$
$16=\frac{1}{2}(10-5\sqrt{3})t^2$
$32=(10-5\sqrt{3})t^2$
$\frac{32}{(10-5\sqrt{3})}=t^2$
$\frac{32(10+5\sqrt{3})}{25} = t^2$ (rationalise)
$t^2 = \frac{16(20+10\sqrt{3})}{25}$
$t=\sqrt{\frac{16(20+10\sqrt{3})}{25}}$
$\therefore t=\frac{4}{5}\sqrt{20+10\sqrt{3}}$
BTW this is more of a physicsy questions, if it was truly 4u it would be resisted.

its in 3u forum lol

notme123 · Aug 30, 2021

idkkdi said:
its in 3u forum lol

oh lol whoops i didnt read.

CM_Tutor · Sep 1, 2021

Resolving forces in this sort of way is definitely part of the vectors material of the new MX1 syllabus, but it will take some years to get a really clear picture of what is considered "fair" for examination.

Force on inclined slope (1 Viewer)

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