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

Answer

A

Explanation

(a) From work-energy theorem,
Work done=Change in KE
`rArr " " W=K_(f)-K_(i)`
`rArr " " K_(f)=W+K_(i)=int_(x_(1))^(x_(2))Fx dx+(1)/(2)mv^(2)`
`=int_(20)^(30)-0.1.xdx+(1)/(2)xx10xx10^(2)`
`-0.1[x^(2)/(2)]_(20)^(30)+500=-0.05[30^(2)-20^(2)]+500`
`=-0.05[900-400]+500`
`rArr " " K_(f)=-25+500=475J`