Initial velocity of the block `vecu=45hati` while external force on it is `vecF=25(-hati)` it coefficient of static and kinetic friction are 0.3 and 0.2 respectively then distance travelled by the block in 12 sec `(g=10m//s^(2))`

Options

A. 225 m
B. 216 m
C. 210 m
D. 250 m

Answer

A

Explanation

Net retardation a `=(F_(ext)+mu_(k)mg)/(m)=4.5`
if then stop at the t, then
`V=u+at`
`0=45-4.5timpliest=10sec`
When block stops, `F_(ext)` will try to bring the block back ward while frictional force will oppose its motion, since block is stationary therefore at this moment, frictional force will be static. Whose maximum value will be `mu_(s)mg=30N`. Since static frictional force is self adjusting therefore it will be 25 N and block will not move after t=10sec.
`S=(u^(2))/(2a)=225m`