Question

An object with a mass of `2 kg` is acted on by two forces. The first is `F_1= < 2 N , -7 N>` and the second is `F_2 = < 1 N, -6 N>`. What is the object's rate and direction of acceleration?

Answer 1

The rate of acceleration is `=6.67ms^-2` in the direction of `=103`º

Explanation:

The resultant force is

`vecF=vecF_1+vecF_2`

`= <2,-7>+<1,-6>`

`=<3,-13>`

We apply Newton's second Law

`vecF=m veca`

Mass, `m=2kg`

`veca =1/m*vecF`

`=1/2<3,-13> = <3/2, -13/2>`

The magnitude of the acceleration is

`||veca||=||<3/2,-13/2>||`

`=sqrt((3/2)^2+(13/2)^2)`

`=sqrt(178)/2=6.67ms^-2`

The direction is `theta = arctan((-13)/(3))`

The angle is in the 2nd quadrant

`theta=arctan(-13/3)=180-77=103`º

Answer 2

`a=sqrt(178)/2" "m/s^2 " , "alpha ~=-89.bar 9 ^o`

Explanation:

`F_1= <2N,-7N >" "rArr" "vec F_1=2i-7j`

`F_2= <1N,-6N >" "rArr" "vec F_2=i-6j`

`vec F_("net")=(2+1)i+(-7-6)j`

`vec F_("net")=3i-13j`

`F_("net")=sqrt(3^2+(-13)^2)`

`F_("net")=sqrt(9+169)`

`F_("net")=sqrt(178)" "N`

`"acceleration : a"`

`"mass :m"`

`a=F_("net")/m" The Newton's second law of motion"`

`a=(sqrt(178))/2" "m/s^2`

`tan alpha=-13/3`

`tan alpha=-4.bar 3`

`alpha ~=-89.bar9^o`