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shivanshi asthana Grade: 11

hey m sorry i cnt post my mail id publicly lik if u cn,kindly giv me.....m from class 11....n ans to that ques is 20j.....hav u considered 5 N force in horizontal direction only???

8 years ago

Answers : (4)

Badiuddin askIITians.ismu Expert
147 Points

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 = Si + 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

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8 years ago
Shreyans Sharma
33 Points

yeah fine ill give you mine..... My E-mail ID is 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

8 years ago
Shreyans Sharma
33 Points

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

8 years ago
Shreyans Sharma
33 Points

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.

8 years ago
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