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hey m sorry i cnt post my mail id publicly lik this..............bt if u cn,kindly giv me.....m from lko...in class 11....n ans to that ques is 20j.....hav u considered 5 N force in horizontal direction only???
Dear shivanshi
given initial velocity = 4i + Oj
final velocity = 0i + 6j
let force is F =li + mj where √l2 +m2 =5
acceleration of the body = F/m =1/2 (li + mj)
or ax = l/2 and ay = m/2
since final velocity in horizontal direction is 0
so horizontal distance moved by container when velocity changes from 4 m/sec to 0
use V2 =u2 +2axsx
0 = 16 + 2*l/2 Sx
Sx =-16/l
and vertical distance moved by container when velocity changes from 0 to 6 m/sec
use V2 =u2 +2aySy
36 = 0 + 2*m/2 * Sy
Sy =36/m
so displacement vector = Sx i + Syj
=-16/l i + 36/m j
now work done W=F.d
=(li + mj ).(-16/l i + 36/m j)
=-16 +36
=20 j
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
yeah fine ill give you mine..... My E-mail ID is shreyans_sharma@hotmail.com. Nd i used the formula for change in kinetic energy using the magnitudes of the velocity given i.e. 4m/s nad 6m/s respectively. As work done = Change in K.E. The method is simple. I had a resonance test today wasnt good though. Please contact me on my ID
Thanks for ur gesture please contact me on my ID i would be waitin for ur message...........and I also am in class XI, Shimla
hey Ill be online upto 2:00 a.m. wenever u get time surely contact me.........I gotta make some online friends now.....n dis is the best place wer i can find sme intellectual friends.... Plzzzzzz contact me waitin fr ur reply.
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