A block of mass 2kg is placed on ground. It is pulled upward with variable force `F= 30(1-2y),` where `y=` height from ground. Assume there is NO air drag NEITHER any other resistive force, then K.E. of the block at half of the maximum height achieved by it is given as :-
`(g =10 m//s^(2))`

Options

A. `1 J`
B. `5/6 J`
C. `4/3J`
D. `3/2 J`

Answer

B

Explanation

by F.B.D of block
`F- mg =ma`
`30 (1- 2y) - 20= 2a`
`a= (10- 60y)/(2) =5- 30y`

`v (dv)/(dy) =5- 30y`
`underset( 0 ) overset(0 ) int vdv = underset(0) overset(H) int (5- 30y) dy`
`0=5H - 15 H^(2)`
`H =1/3 m`
Now speed at half maximum height
`int vdv = overset( 1//6) underset(0) int (5-30) dy`
`(v ^(2))/(2) =5 ((1)/(5))-15 ((1)/(6))^(2)`
`(v ^(2))/( 2) = (5)/(12)`
`KE=1/2 mv ^(2) =5/6 J`