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  a_x = l/2    and a_y = 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 +2a_xs_x

       0 = 16 + 2*l/2 S_x

 S_x =-16/l

and vertical  distance moved by container when velocity changes from 0 to 6 m/sec

use  V² =u² +2a_yS_y

       36 = 0 + 2*m/2 * S_y

  S_y =36/m

so displacement vector = S_x  i + S_yj

                                 =-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 Experts
Badiuddin

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.