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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 May 2008 17:14:19 IST
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a load of mass 10kg is raised through a height of 15 m frm the ground where it ws initially at rest. when it is at that height it has a velocity of 20 m/s .find the work done....
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Memory,the daughter of attention, is the teeming mother of knowledge!!!! |
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3 May 2008 17:20:43 IST
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v2 - u2 =2as
therefore a= 40/3
force = m*a =10*40/3 =400/3
WD= F.s = (400/3)*15 = 2000 J
RATE IF USEFUL
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3 May 2008 17:34:30 IST
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Initial ME + work done = Final ME
m g (0)+1/2 m (0) + W= m g (h) + 1/2 m (400)
substituting values.....
W= 1500+ 2000 = 3500J , taking g=10 SI
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3 May 2008 17:37:11 IST
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work done is change in kinetic energy
so answer is 2000J
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3 May 2008 18:07:05 IST
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W- work done
now
W + mgh = 1/2mv^2
W =1/2 ( 10) (20)^2 - 10*10*15
= 2000 - 1500
= 500J
so therefore the work done = 500J
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4 May 2008 07:16:52 IST
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buit the answer is 2190j .....
nd the problem which i think here is that at the height it has also some velocity ....
some other answer plzzzzz
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4 May 2008 07:31:28 IST
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the answer is 3500 J
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4 May 2008 11:48:35 IST
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Well I have some other approach but answer as 3500Joules see how.......
Suppose at the height of 15m the body was left free....but as it has some velocity i.e. 20m/s at that instant...then it will continue its upward motion for some distance and attain a maximum height....
and thus Potential Energy of the Body at that instant will be the Total Work Done on the Body.....
so.....using
we have....
using g = 10.......gives h = 20 m...
so final height = 15 + 20 = 35 m.......
and Potential Energy at 35m from ground = mgh = 3500 Joules........
If somebody has some other approach then pls do post.........
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4 May 2008 13:40:22 IST
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i agree with aatish
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4 May 2008 14:22:31 IST
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ur approach is gud aatish
bt wat i hav done is simply
Work done= Change in P.E. + Change in K.E.
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5 May 2008 18:27:59 IST
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total work done by all forces = changein kinetic energy = 2000J
now there is an ambiguity as to whether the ques asks for totalwork done or work done by external force
so work done by gravity = mgh = -1500J
now total work done = work done by gravity + work done by external force
so work done by external force = 2000 + 1500 = 3500
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