A block of mass 10 kg, initially at rest, makes a downward motion on `45^(@)` inclined plane. Then the distance travelled by the block after 2s is
( Assume the coefficient of kinetic friction to be 0.3 and `g=10ms^(-2)`)

Options

A. `7sqrt(2)m`
B. `(9)/(sqrt(2))m`
C. `10sqrt(2)m`
D. `5sqrt(2)m`

Answer

A

Explanation

`v=u+at`
`rArr 8 hat(i)+10 hat(j)=3 hat(i)+axx8`
`rArr a=1/8 (5 hat(i)+10hat(j))m//s^(2)`
So, force `F=ma`
`F=4/8(5 hat(i)+10hat(j))=1/2(5hat(i)+10hat(j))N`
`rArr F=1/2sqrt(5^(2)+10^(2))=(5sqrt(5))/(2)N`
Hence, the option (a) is correct.