A body of mass `10kg` is acted upon by two perpendicular force, `6N` . The resultant acceleration of the body is.

Options

A. `1ms^(-2)` at an angle of `tan^(-1)(3/4)` w.r.t. `8N` force
B. `0.2ms^(-2)` at an angle of `tan^(-1)(3/4)` w.r.t. `8N` force
C. `1ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t. `8N` force
D. `0.2ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t. `8N` force

Answer

A

Explanation

Here, `m=10kg`
The resultant force acting on the body is
`f=sqrt((8N)^(2)+(6N)^(2))=10N`
Let the resultant force `F` makes an angle 6 w.r.t. `8N` force.
From figure, `tantheta=(6N)/(8N)=(3)/(4)`
The resultant acceleration of the body is
`a=(F)/(m)=(10N)/(10kg)=1ms^(-2)`
The resultant acceleration is along the direction of the resultant force.
Hence, the resultant acceleration of the body is `1ms^(-2)` at an angle of `tan^(-1)` w.r.t. `8N` force.