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 ms ^(-2)`
B. `8.0 ms ^(-2)`
C. `3.17 ms ^(-2)`
D. `10.0 ms ^(-2)`

Answer

A

Explanation

FBD of the block is shown below

Kinetic friction acting on the body
`= mu _(k) R = 0.2 (Mg - F sin 30^(@))`
`= 0.2 ( 5 xx 10- 40 xx (1)/(2)) = 0`
`2 (50 - 20) = 6 N`
By Newton.s second law of motion. acceleration of the block
`("net force")/("mass") = (F cos 30^(@)"- kinetic friction")/("mass")= ( 40 xx (sqrt3)/(2) - 6)/(5) = 5. 73 m//s ^(2)`