A body moving at 2 m/s can be stopped over a distance x.If its kinetic energy is doubled,how long will it go before coming to rest,if the retarding force remains unchanged?
(a) x (b) 2x (c) 4x (d) 8x

Dear Student,
Please find below the solution to the asked query:

Let the mass of the body be 'm'. then,KE of the body = KE= 12mv2 = 12m×22 = 2mNow,KEnew = 2×KE = 4m4m = 12mv'2v' = 22 v --- (1)Now, In the first case when the body is stopped over a distance 'x' applying the third equaitonof motion:0 = 22 - 2axa = 2x --- (2)since the retarding force remains the same in the second case as well then, if the body stops after a distnace x' then,from the third equatio of motion:v2 = u2 -2ax'02 = v'2 -2ax'   8 = 2×2x×x'   (substituing the values from eq1 and eq2)x' =2xTherefore the body will now stop at a distance of '2x'. Thus the correct option is (b)

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