# A question about speed [closed]

A object with mass 0.5 kg slides in straight line for 200 m, where the friction act on it is 0.4 times the normal reaction acting on it. Find its initial speed.

I let $u$ be the speed.

$u^2=2as=2*0.4mg*200=2*0.4*0.5*9.81*200,u=28.01ms^{-1}$.

But the answer should be $40ms^{-1}$, can someone tell me what's wrong with my solution?

Thank you.

EDIT:

$u^2=-2as=-2(f/m)s=-2(-0.4N/m)s=-2(0.4mg/m)s=2*0.4*10*200=1600$

$u=40ms^{-1}$

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## closed as too localized by Nathaniel, dmckee♦Jun 22 '13 at 16:59

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Can you extend your explanation of what you did? I can't follow your steps. –  Bernhard Jun 22 '13 at 8:39
After which = sign can't you follow? –  ᴊ ᴀ s ᴏ ɴ Jun 22 '13 at 8:41
Are you sure you've quoted the question correctly? –  David M. R. Jun 22 '13 at 9:35
Great work! But how did you just forget about the negative sign in the second last step? The friction is in the opposite direction, and so we take it to be -(0.4/m), not positive! That's how the two negatives give a positive - you've just neglected the negative after the third last step, which is wrong. –  mikhailcazi Jun 22 '13 at 12:59
Done. :) ${}{}{}{}{}$ –  ᴊ ᴀ s ᴏ ɴ Jun 23 '13 at 3:10

You're solution is wrong. For one, the law is

$v^2 = u^2 + 2aS$

where v is the final velocity and u is the initial velocity.

You've substituted the initial velocity in the wrong place, can't you see? :)

Secondly, why do you think the acceleration is equal to 0.4(N)? It is given in the question that the friction is equal to 0.4(N), not acceleration! Think of a way to inter-relate the two?? ;)

Thirdly, try taking the acceleration due to gravity (g) as 10, it'll simplify stuff.