Two 20-g flatworms climb over a very thin wall, 10 cm high. one of the worms is20 cm long the other is wider and only 10 cm long. Which of them has done more work against gravity when half of it is over the top of the wall? What is the ratio of the amounts of work done by the two worms?
21
first of all sorry for the ridiculous image......[4]
now....we see that when longer worm is half way over the wall its centre of mass is at a height of 5cm and when shorter worm is half-way over the wall its centre of mass is at height 7.5 cm.
since mass of both worms are same.....therefore mass of half of both worms would be same too. [3]
work done in 1st case=10*5*g
work done in 2nd case=10*7.5*g.
so ratio is 2:3.
1wall is vry thin
the earthworm will go down on other side
its a bt ques
worm is nt on ne thick wall so will come dwn on other side
it will make a "U" shape opening dwnwrd
49oops din see ur posts as i was posting!! neways i give my views for the question!!
the work is done by the insect things against gravity upto the time when their body become straight to full length of the wall!! 2nd case in both images!!
after that the work done by the insect against gravity is balanced by the work done by gravity on insect!!
thus work done= displacement of the CoM * m * g
for first case: 2.5*20*980
for second case: 5*20*980
verify if i am right or wrong!!
21@nikunj even if the wall is thin will it change the Y- coordinates of centre of mass? it will still be at 5cm and 7.5cm respectively. the only thing that will be changed is mass.....as now we will have to take whole mass instead of half of mass.........but it wont affect the ratio of work done.
49yea it will coz the centre of mass will not move after the situation in fig 2 and fig 5 in #6 is reached...think on this!!!
my solution for the first one is looking to be wrong to me!! ruko i ll check!!!
49sorry the concept i used now loos to be wrong to me!!
work=y displ of CoM X m X g
using this, fro case 1 it is : (2.5 +7.52)X20X980
case 2: (5+52)X20X980
