Question

An object with a mass of `3 kg` is acted on by two forces. The first is `F_1= < 2 N , 8 N>` and the second is `F_2 = < 2 N, -3 N>`. What is the object's rate and direction of acceleration?

Answer 1

The rateof acceleration is `=2.13ms^-2` in the direction `=51.3`º

Explanation:

The resultant force is

`vecF=vecF_1+vecF_2`

`= <2,8>+<2,-3>`

`=<4,5>`

We apply Newton's second Law

`vecF=m veca`

Mass, `m=3kg`

`veca =1/m*vecF`

`=1/3<4,5> = <4/3, 5/3>`

The magnitude of the acceleration is

`||veca||=||<4/3,5/3>||`

`=sqrt((4/3)^2+(5/3)^2)`

`=sqrt(41)/3=2.13ms^-2`

The direction is `theta = arctan((5)/(4))`

The angle is in the 1st quadrant

`theta=arctan(5/4)=51.3`º

Answer 2

We are given two forces
`F_1=2hati+8hatj`
`F_2=2hati-3hatj`
`:.`Net force `F= F_1+F_2`
`=>F=4hati+5hatj`
Now by Newton's second law
`F=mxxa`
`=>4hati+5hatj=3xxa`
`=>a=1/3 (4hati+5hatj)`