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

From work-energy theorem,
Work done = Change in KE
`implies W=k_(f)-K_(i)`
`implies K_(f)=W+K_(i)=int_(x_(1))^(x_(2)) Fx dx+1/2 mv^(2)`
`=int_(20)^(20) -0.1.x dx+1/2 xx10xx10^(2)`
`=-0.1 [x^(2)/2]_(20)^(30)+500`
`=-0.005 [30^(2)-20^(2)]+500`
`=-0.05 [900-400]+500`
`implies K_(f)=-25+500=475 J`