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lionel messi · Sep 16, 2012, 09:30 AM

An object weighing 98N is dropped from a height of 10m. It is found to be moving with a velocity 12m/s just before it hits the ground. How large was the frictional force acting upon it?
Will it be done by using k.E=P.E+W in which -26N ANSWER is coming. If yes,than can it be solved via equations of motion (than f=ma)..

Unknown008 · Sep 16, 2012, 09:45 AM

Both methods can be used.

lionel messi · Sep 16, 2012, 09:52 AM

But answers are different?? I think 26N is the right answer which I am getting from work energy equation. But can't from equ of motion?

Unknown008 · Sep 16, 2012, 10:03 AM

Show your work of how you reached 26 N please.

lionel messi · Sep 16, 2012, 10:08 AM

0.5mv^2-mgh=F.S
0.5*10*12^2 - (10*9.8*10)=f.s
-260N=F(10)
F=-26N..

Unknown008 · Sep 16, 2012, 10:09 AM

Okay, how did you do for the equation of motion?

lionel messi · Sep 16, 2012, 10:13 AM

I am getting different answers.but not 26N.. Can u solve it please..

Unknown008 · Sep 16, 2012, 10:14 AM

No no, I want you to post what you tried. There's nothing wrong with making a mistake. You learn from mistakes after all.

lionel messi · Sep 16, 2012, 10:20 AM

Here vi=0 or vf=0..

Unknown008 · Sep 16, 2012, 10:22 AM

I won't answer anything more until you post what you tried, from the start to the final answer you got that you're saying is wrong.

lionel messi · Sep 16, 2012, 10:27 AM

Ive delivered my try before as well..
If vi=0 than,
f=ma
=10*(vf-vi)/t
=10*(12/1.428)
=84N.. WHERE t from 2nd equ of motion

Unknown008 · Sep 16, 2012, 10:31 AM

Okay, I'm not sure why you tried to use something you have to calculate here.

I would use $v^2 = u^2 + 2as$

v = 12
u = 0
a = ?
s = 10

$(12)^2 = 0^2 + 2a(10)$

$a = 7.2 m/s^2$

This is the net acceleration of the object.

$a_{net} = a_g - a_f$

$7.2 = 9.8 - a_f$

$a_f = 2.6 m/s^2$

This is the acceleration due to friction.

F = ma

F = 10*2.6 = 26 N

lionel messi · Sep 16, 2012, 10:35 AM

Thanks mate..