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Maths/Physics help

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astarmathsandphysics:

--- Quote from: omarsubei on February 12, 2009, 10:14:17 am ---
--- Quote from: astarmathsandphysics on February 12, 2009, 08:38:02 am ---
--- Quote from: omarsubei on February 11, 2009, 11:45:41 pm ---Please help me with this question and explain to me why the answer is what it is:

A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:

A. 2.
B. 4.
C. 8.
D. 16.

Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4


--- End quote ---

--- End quote ---

That's the answer I came up with, but strangely the mark scheme says the answer is C (8). I was thinking maybe because it's asking about the "Work" done by F, and that there is distance involved (W=Fd), that the distance in F2 is double that of F1 because it's twice as fast?

--- End quote ---
0.5(2M)V^2/0.mvv^"=8 THEquestion is badlty worded

omarsubei:
Sorry I don't understand. Could you please explain.

astarmathsandphysics:
f2 is not twicw F1

astarmathsandphysics:
u hve to work out the ratiof2/f1 first.

omarsubei:
You're making me feel even more confused. :-\
The ratio of F2/F1 is 4. How does the "work"  make that 8?

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