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M2 Work done question: Where did I go wrong?

The mark scheme did the whole "change in energy = wd by friction" which I understand but I don't understand how my method was incorrect..

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Reply 1
Original post by creativebuzz
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You're going to hit yourself. :tongue:

The weight is 0.6g0.6g so your equation should be 0.6gsin30Fr=0.6×230.6g\sin 30^{\circ} - F_r = 0.6 \times \frac{2}{3} - then solve for FrF_r.

That took me far too long to spot. :lol:
Original post by Zacken
You're going to hit yourself. :tongue:

The weight is 0.6g0.6g so your equation should be 0.6gsin30Fr=0.6×230.6g\sin 30^{\circ} - F_r = 0.6 \times \frac{2}{3} - then solve for FrF_r.

That took me far too long to spot. :lol:


Hahah! Oh god that's so stupid of me!

But.. I still get 88.9N which isn't the right answer :/
Reply 3
Original post by creativebuzz
Hahah! Oh god that's so stupid of me!

But.. I still get 88.9N which isn't the right answer :/


How do you get that? Re-arranging: Fr=0.6×9.8×120.4F_r = 0.6 \times 9.8 \times \frac{1}{2} - 0.4 - so what do you get for FrF_r?
Original post by Zacken
How do you get that? Re-arranging: Fr=0.6×9.8×120.4F_r = 0.6 \times 9.8 \times \frac{1}{2} - 0.4 - so what do you get for FrF_r?



Ohhh because I divided!!! *facepalm**

Thank Zacken! I was freaking out because I thought my method was completely wrong but I didn't know why!
Reply 5
Original post by creativebuzz
Ohhh because I divided!!! *facepalm**

Thank Zacken! I was freaking out because I thought my method was completely wrong but I didn't know why!


Nope your method was perfectly fine, well done.

No problem.
Original post by Zacken
Nope your method was perfectly fine, well done.

No problem.


Such a relief!

Also,

do you mind explaining why the reaction forces act left and down? I'm still struggling to understand this and it's been soo long

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Reply 7
Original post by creativebuzz
Such a relief!


Looks like Ghostwalker's going to answer that one, I'll let him do the honours. :-)

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