NEET Class neet Answered

Intial speed of particle at x= 0 is v_o = 4 m/s

speed v₄ of particle at x =4  is calculated from equation of motion " v₄² = v_o² + ( 2 a S) " where a = 2 m/s² is uniform accelerartion for the initial distance S = 4 m .

from x = 4 m to x = 8 m , acceleration changes with position

Let a = p x + q

at x =4 , a = 2 ; hence  4p+q = 2

at x =8 , a = 0 ; hence  8p+q = 0

By solving above equations , we get p = - 0.5 and q = 4 .

hence acceleation is

By integrating above expression on both sides from x = 4 m to x = 8 m , we get

Hence , we get speed v₈ at x= 8 m by substituting v₄ =   m/s as  , v₈ =   m/s

Hence . final kinetic energy is

---------------------------------------------------------------------------

By work-energy theorem, chane in kinetic energy equals Work done on partcile

Initial kinetic energy of particle at x = 0 m is 

Change in kinetic energy = ( 20-8) J = 12 J

Total workdone on the particle = 12 J

-----------------------------------------

Workdone by conservative force = loss of potential energy = 120 - (-120)  J  = 240 J

-----------------------------------------------

Total workdone  = Workdone by external force +  workdone by conservative force

Workdone by external force = ( 12 - 240  )  J = -228 J

--------------------------------------------

My answers are not matching with the options in the given list .

However, I have given the correct method of calculating final energy , Total workdone . workdone by conservative force

and workdone by external force .