A 1.5 kg block is kept stationary on a horizontal surface.A force f= 4 - x2 starts acting on it on the COM in the horizontal direction.the max K.E of the block between 0<=x<=2m is(where <= denotes greater than or equal to)
a)2.33
b)6.87
c)5.33
d)6.67
ans-c
-
UP 0 DOWN 0 0 2
2 Answers
1we see that acceleration is positive till x=2 .
thus velocity of the block is increasing till x=2 in +ve direction.
thus in 0<x≤2 max. K.E is at x=2 .
given
ma=4-x2
=> mv.dvdx= 4-x2
=> m∫v.dv = ∫ (4-x2 ) dx
=> Mv22 = 4x-x33
we observe L.H.S is K.E
putiing x=2.
KE= 16/3=5.33
24just another way of writing what anirudh did..
applying WE theorm,
ΔK=ΔW
2
ΔK=∫F.dx
0
ΔK=16/3
