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

Options

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

Answer

C

Explanation

Here, `10kg, v_(i)=10ms^(-1)`
Initial kinetic energy of the block is
`K_(i)=(1)/(2)mv_(i)^(2)=(1)/(2)xx(10kg)xx(10ms^(-1))^(@)=500J`
Work done by retarding force
`W=int_(x_(1))^(x_(2))F_(r)dx=int_(20)^(30)-0.1xdx=-0.1[(x^(2))/(2)]_(20)^(30)`
`=-0.1[(800-400)/(2)]=-25J`
According to work - energy theorem, `W=K_(f)-K_(i)`
`K_(f)=W+K_(i)=-25J+500J=475J`