A block of mass `m=5kg` is resting on a rough horizontal surface for which the coefficient of friction is `0.2` . When a force `F=40N` is applied, the acceleration of the block will be `(g=10m//s^(2))` .

Options

A. `5.73m//sec^(2)`
B. `8.0m//sec^(2)`
C. `3.17m//sec^(2)`
D. `10.0m//sec^(2)`

Answer

A

Explanation

Kinetic friction `=mu_(k)R=0.2(mg-Fsin30^(@))`
`=0.2(5xx10-40xx(1)/(2))= 0.2(50-20)=6N`
Acceleration of the block
`=(Fcos30^(@)-Ki n etic f riction)/(Mass)`
`=(40xxsqrt(3)/(2)-6)/(5)=5.73m//s^(2)` .