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the position x of a particle moving along xaxis at time t is given by equation t=√x +2.find the work done by the force in first 4seconds the position x of a particle moving along xaxis at time t is given by equation t=√x +2.find the work done by the force in first 4seconds
Hi Vaibhav,Again it is calculusafter 4seconds, S =(4-2)2=4 W = F*s = m*a*s = m*2*4 =8m (you have not given mass of th particle) Plz approve If you are satisfied :)(previous question also :P)Let me know if you still have doubts..........
But mr. strange mass is not given in the question and the right answer is 0 J,,,, i dont know how but its d right answer,,,, and please view the prev. question and help me out in that too please.... and thnx for ...answering mine doubts... ;)
sry I madv=2(t-2)initial velocity ( at time t =0) u = 2(0-2) = -4s = u*t + 1/2*a*t2 after 4seconds , s= -4*4 + 1/2*2*42 = -16 + 16 = 0s=0, hence NO work done , W=0JAnd I have cleared your doubt of previous que plz check :)
Here , we have been given an equation t=√x + 2.Let's find X. √X = t-2X = t^2-4t+4Now let's find velocity.dx/dt = 2t-4.Put dx/dt = 0.By this way , we get the time at which the particle's velocity is zero. We get t =2 sec.Now in the question We've been asked to find the work done in t= 4 seconds . THIS CLEARLY MEANS THAT THE PARTICLE IS REVERSING ITS DIRECTION. Hence at 4 seconds the displacement of the particle is zero . Hence work done is zero.
Dear Student,Please find below the solution to your problem.v=2(t-2)initial velocity ( at time t =0) u = 2(0-2) = -4s = u*t + 1/2*a*t2after 4seconds , s= -4*4 + 1/2*2*42= -16 + 16 = 0s=0, hence NO work done , W=0JThanks and Regards
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