It would be great if you can help me
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Rough outline:It would be great if you can help me
in ext2 mechanics do we usually just assume v is always greater than zero even if velocity is in the downward direction?Rough outline:
View attachment 36664
For (i), draw up a diagram of the mass, with the forces and take the resistive forces (going up the slope) as negative and the weight force parallel to the mass (mgsintheta) as positive:
So net force would be mgsinx - 2v - 2v^2
and sub m =5, theta = 60 degrees and you'll get the required answer.
(ii)
For constant speed, a = dv/dt (rate of change of velocity) = 0 i.e net force = 0 as f = ma:
So you'll get a quadratic in (i) and use the quad formula to solve for v.
For a guide, times equation inn (i) by 2 to get
v > 0 so only consider positive
Simplify or use your calculator and you should get 4.2 ms^-1.
so v = 4.2 ms^-1
Also the solutions should be on the HSC marking criteria on the website?
You can take the downwards direction as positive in that scenarioin ext2 mechanics do we usually just assume v is always greater than zero even if velocity is in the downward direction?
Aight thanks bro. Any other advice/foresight for us tomorrow for the exam? Like any silly mistakes you made or some stuff you didn't know after the exam? CheersYes and make sure if it asks to 1dp, put 1dp. They WILL take marks off for not doing so. Seems trivial but you don't want to lose a mark on something like that so read the q
I just stated in my solutions that I am taking downward as positive just because but I presume that they assume in that context that downward direction is positive given than the forces are considered "resistive".in ext2 mechanics do we usually just assume v is always greater than zero even if velocity is in the downward direction?
Keep your focus.Aight thanks bro. Any other advice/foresight for us tomorrow for the exam? Like any silly mistakes you made or some stuff you didn't know after the exam? Cheers