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friends i could post only half the experiment..in the previos post.... the rest is below
2 Answers
now how long will it take for the entire length of chain
to run down like a thread ??. as per calculations ,(taking
acceleration due to gravity as 10 metres per sec per sec .and there
is zero air resistance .the process is considered as
frictionless as possible.) the time it will take for 80 metres
of chain to run down and hit the ground will be 11.315
seconds . and at the piont of hitting the ground every part of
the chain will have a velocity of 7.07 metres per second. so
total K.E OBTAINED when 80 metres of chain rundown will be
K.E = 1/2 *M*V*V
= 1/2 *80*7.07*7.07
= 2000 UNITS .
WHICH IS half the energy required to lift 80 metres of chain
to a height of 5 metres as given below.here 7.07metres per
sec is the terminal velocity when height is 5 metres.{when
height is 20m term velocity is 14.14 m per sec}.
here law of conservation of momentum works but not L.o.C.E..
((when u drop all of the chain together it will hit
the ground in 1 seconds ( take acceleration due to gravity as
10 m per sec per sec). at time of hiting the ground every
part of chain will have a velocity of 10m per second.this way
P.E = K.E .(P.E = M*G*H, K.E 1/2 *M*V*V)
I.E., P.E = K.E.
80*10*5 = 1/2*80*10*10.
4000 units = 4000 units.))
THIS EXP CAN BE DONE USING MUCH LONGER LENGTHS OF CHAINS.
THE HEIGHT FROM WHICH U DROP IT CAN ALSO BE VARIED ..
my contact address is [edited by AB ]
to run down like a thread ??. as per calculations ,(taking
acceleration due to gravity as 10 metres per sec per sec .and there
is zero air resistance .the process is considered as
frictionless as possible.) the time it will take for 80 metres
of chain to run down and hit the ground will be 11.315
seconds . and at the piont of hitting the ground every part of
the chain will have a velocity of 7.07 metres per second. so
total K.E OBTAINED when 80 metres of chain rundown will be
K.E = 1/2 *M*V*V
= 1/2 *80*7.07*7.07
= 2000 UNITS .
WHICH IS half the energy required to lift 80 metres of chain
to a height of 5 metres as given below.here 7.07metres per
sec is the terminal velocity when height is 5 metres.{when
height is 20m term velocity is 14.14 m per sec}.
here law of conservation of momentum works but not L.o.C.E..
((when u drop all of the chain together it will hit
the ground in 1 seconds ( take acceleration due to gravity as
10 m per sec per sec). at time of hiting the ground every
part of chain will have a velocity of 10m per second.this way
P.E = K.E .(P.E = M*G*H, K.E 1/2 *M*V*V)
I.E., P.E = K.E.
80*10*5 = 1/2*80*10*10.
4000 units = 4000 units.))
THIS EXP CAN BE DONE USING MUCH LONGER LENGTHS OF CHAINS.
THE HEIGHT FROM WHICH U DROP IT CAN ALSO BE VARIED ..
my contact address is [edited by AB ]
Answers
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I'm sorry if you have spent 4 years on this as far as I can tell you are saying that dropping all the chain at once gives different figures than letting it "flow" over the edge taking 11 seconds. As above you cannot ignore friction, between the links and between the chain and the surface it is resting on, there is also the energy required to overcome the links that are stationary, waiting to be pulled over the edge. This is not a LCE experiment at all, it's more ove an ADTG experiment.
Are you a student? Go back to your tutor with this I'm sure he/she can explain.
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