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

Answer

A

Explanation

Apply work-energy throrem.
When a force acts upon a moving body, then the kinetic energy of the body increases and the increase is equal to the work done. This is work energy theorem.
Work done `= (1)/(2)mv^2 - (1)/(2) m u^2 = K_f - K_i`
Another definition of work done is force `xx` displacement
`:. F dx = K_f - (1)/(2) mv_i^2`
where the subscripts `f` and `i` stand for final and initial
`F.dx = K_f - (1)/(2) xx 10 xx (10)^2`
`rArr F .dx = K_f - 500`
`rArr underset(x =20)overset(x = 30)(int) (-0.1)x dx = 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`
`rArr K_f - 500 = -25`
`rArr K_f = 500 - 25 = 475 J`.