A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .

Options

A. `250 J`
B. `450 J`
C. `275 J`
D. `475 J`

Explanation

When a force acts upon a moving body , then the kinekic energy of the body increase and the increase is equal to the work done. This is work energy therom.
Work done `= (1)/(2) mv^(2) - (1)/(2) mu^(2) = K_(f) - K_(i)`
Another definition of work is force xx displacement.
`:. Fdx = K_(f) - (1)/(2) xx mv_(1)^(2)`
where the subscripts `f` and l` stand for final and initial.
`F. dx = K_(f) - (1)/(2) xx10 xx (10)^(2)`
`implies F. dx = K_(f) - 500`
`rArr underset(x=20)overset(x=30)int=K_(f)-500`
Using the formula `int x^(n) dx = (x^(n+1))/(n+1)`, we have
`- 0.1 [(x^(2))/(2)]_(x=20)^(x=30) = K_(f) - 500`
`- 0.1 [(30)^(2)/(2) - ((20)^(2))/(2)] = K_(f) - 500`
`implies K_(f) - 500 = - 25`
`implies K_(f) - 500 - 25 = 475J`