In rolling without sliding on a stationary ground, work done by friction is zero. Hence
work done by the applied force
`=` change in kinetic energy
`therefore(30)(0.25)=(1)/(2)xx9xxv^(2)`
`+2[(1)/(2)xx6xxv^(2)+(1)/(2)xx(1)/(2)xx6xxr^(2)xx(v^(2))/(r^(2))]`
or `7.5=13.5v^(2)`
`thereforev=0.745m//s`