A block of mass 20 kg is moving in x-direction with a constant speed of 10 `ms^(-1)`. It is subjected to a retarding force `F=(-0.1x)N` during its travel from x=20 m to x=30 m. Its final kinetic energy will be

Options

A. 475 joule
B. 450 joule
C. 275 joule
D. 250 joule

Answer

A

Explanation

According to work energy theorem,
`W=Delta K.E. rArr W=(1)/(2) mv_(2)^(2)-(1)/(2)mv_(1)^(2)`
`rArr" "F cdot dx=K.E._(F)-(1)/(2)xx10xx100`
`rArr F cdot dx=K. K._(F)-500`
`rArr overset(30)underset(20)-0.1 cdot dx =K.E._(F)-500`
`rArr" "-0.1[(x^(2))/(2)]_(20)^(30)=K.E._(F)-500`
`rArr" "-0.1[(900-400)/(2)]=K.E._(F)-500`
`rArr" "K.E._(F)=500-25=475 J`.