Need help past HSC question (1 Viewer)

[lm199]

2007 HSC question 13 c (ii)

my teacher gave a different answer to one of the answers from a "possible answers" pdf file

thanks

 

[Riproot]

Can you post the question here?

Just screen-shot it, paint it, and upload to here.

Cbf looking it up.

 

[lm199]

 

[lm199]

i got 18kN for i) which im pretty sure is right but i have trouble with angles..... one answer says 5.74m and my teacher got 28.11m

 

[lm199]

bump....

 

[barbernator]

i got 18kN for i) which im pretty sure is right but i have trouble with angles..... one answer says 5.74m and my teacher got 28.11m

i got what your teacher got.

 

[lm199]

ahh k thanks

 

[bggy]

I know its been over a year since the last post... but can someone explain how you got that?

 

[nick487]

My understanding of this question is that the "possible solutions" have not included ∆PE, and also, i have no clue where they got there value for mass from. If you include ∆PE, and use the correct mass, you get 28.11m.
the mass can be found as we know the resistance to motion (Frictional force) is µN. And we also know µ=tan5. Therefore 18000=tan5 x mgcos5, and u get m=20.653 tonnes.
now, we let ∆PE + ∆KE =Fs.
F=18000+mgsin5, and now the only trick is to find the height when it stops, but this is found using sin as sin5=h/s, so h=s x sin5.
Then it is a matter of substitution and solving for s.
P.s, i'm not 100% sure this is right, but this is how to get the answer 28.11, or at least how i got it.