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

Answer

A

Explanation

Kinetic friction `=mu_kR=0.2(mg=Fsin30^@)`
`=0.2(5xx10-40xx(1)/(2))=0.2(50-20)=6N`
Acceleration of the block
`=(Fcos30^@-ki n etic f rictio n)/(mass)`
`=(40xxsqrt(3)/(2)-6)/(5)=5.73(m)/(s^2)`