23Taking basket as the origin .......
COM is at a distance of ML/(m + M) frm basket
now since the system is in equilibrium i.e net external force = 0
so displacement of COM= 0
now when the man will finally reach the basket ....
the man and the basket both will hav coordinates of the COM .....
so displacement of basket wrt grnd = ML/(m + M)
now considering the basket as system ..the external agents that can do work on it are man and gravity
so Wgravity + Wman = [change in KE]basket
mgML/(m + M) + Wman = 0
so Wman = - mgML/(m + M)
i kno its gona b wrong [2]
23
qwerty
·2009-10-22 11:50:02
ohh....
thank you sir ....vry good question ..... abhi yaad aya ki wat is bouyant force.....almost forgot it ....
66The important fact, which almost all of you seem to be overlooking, is the fact the balloon + man system is in equilibrium in air. That means that net downward force of gravity must be balanced by some upward force (which comes from the buoyant force on the balloon) F.
So F = (M+m)g.
Since the potential energy of the system does not change, the work done by the man is the negative of the work done by F as the man climbs to the basket. The displacement of the balloon is downward and equals mM+m L in magnitude. So the work done by the force F is
- F mM+m L = -mgL
It is negative since F is up while displacement is down. So the work done by the man = mgL.
23
qwerty
·2009-10-21 22:37:33
sir where are we wrong??? [ deepak and me ?? [4] ]
19i cannot figure out the situation....a figure might be helpful...can anyone provide that?
23
qwerty
·2009-10-21 07:26:33
mera ans thoda alag hai [3] [3] [3] [3]
21yes its wrong ......kaymant sir has already pointed out....[3][3][3][3]
66
kaymant
·2009-10-20 23:49:08
Net work done by the man is not zero.
11Since,net force=0,the centre of mass of the system remains unchanged.Therefore net work done =0
21
eragon24 _Retired
·2009-10-20 22:45:59
hmm........but sir can u plz point out where i m going wrong......i guess we hav to do something like this only......
66
kaymant
·2009-10-20 22:24:38
yes, that answer is wrong.
21
eragon24 _Retired
·2009-10-20 21:51:00
let h be the distance by which the ballon descends when the man reaches upto the ballon by climbing length L
since in verticle direction no external force is acting so com shud be at rest in the verticle direction
so from this we get (m+M)h=mL
mL/(M+m)=h
Gain in pe of man=mgL-mgh
so pe of man increases by mg(L-h)...........and this gain in PE sud be the work done by man to climb the balloon.....
so the work done by man to climb the ballon is mg(L-(mL/(M+m))=MmgL/(M+m)
I know this answer is wrong........ :(
1sir,the work done by man should be 0..as the net force acting on him in y direction is 0...so when he has to climb he has to exert an extra normal force ,but the normal force will only IMPART him an instanteneous velocity and till the normal force acts, the point doesnt displace..........
