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

Options

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

Answer

`(3)`

Explanation

Her, m = 10 kg, ` v_(i) = 10ms^(-1)` Initial kinetic energy of the block is
` K_(i) = 1/2 (mv_(i)^(2)) = 1/2(10kg) xx (10 ms^(-1))^(2) = 500J`
Work done by retarding force
`W = int _(x_(1))^(x_(2)) F_(r)dx = int _(20)^(30) - 0.1x dx = - 0.1[(x^(2))/(2)]_(20)^(30)`
`-0.1[(900 - 400)/(2)] = -25 J`
According to work -energy theorem,
`W = K_(f) - K_(i)`
`K_(f) = W + K_(i) = - 25 J + 500J = 475 J`