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 .

Explanation

According to work-energy theorem,
work done =change in kinetic energy of the vehicle
`:. W=K_(f)-K_(i) orF.dx=K_(f)-(1)/(2)mv_(i)^(2)`
or `F.dx=k_(f)-(1)/(2)xx10xx(10)^(2)or F.dx =k_(f)-500`
or `int_(x=20)^(x=30)(-0.1)x dx=k_(f)-500`
or `-0.1[(x^(2))/(2)]_(x=20)^(x=30)=K_(f)-500`
or `-0.1[((30)^(2))/(2)-((20)^(2))/(2)]=K_(f)-500`
or `K_(f)-500=-0.1(450-200)`
or `K_(f)-500=-25`
`:. K_(f)=500-25=475 J`