# An object with a mass of #14 kg# is acted on by two forces. The first is #F_1= < 5 N , 3 N># and the second is #F_2 = < 3 N, 7 N>#. What is the object's rate and direction of acceleration?

##### 1 Answer

#### Explanation:

We can make use of Newton's second law to solve this problem:

#vecF_(n e t)=mveca# where

#m# is the object's mass and#a# is the object's acceleration

We are given *net* force acting on the object, which is the resultant force caused by

#vecF_(n e t)=vecF_1+vecF_2#

Since our vectors are already broken up into components, we can do simple vector addition to find

#vecF_(n e t)=<5N,3N> + < 3N, 7N>#

#=<8N,10N>#

This is the net force expressed as a vector showing both its parallel and perpendicular components. We will combine them to get a single value (find the magnitude).

#F_(n e t)=sqrt((F_x)^2+(F_y)^2)#

#=sqrt((8N)^2+(10N)^2)#

#=sqrt(164)# #N#

#=2sqrt(41)# #N#

Now we can use the second law to calculate the acceleration.

#veca=vecF/m#

#=(2sqrt(41)N)/(14kg)#

#=0.91m/s^2#

We can find the direction (angle) of the net force using basic trigonometry.

#tan(theta)=(F_y)/(F_x)#

#=>theta=arctan((F_y)/(F_x))#

#=arctan(10/8)#

#=51.3^o# (above the horizontal/positive x-axis)

The acceleration occurs in the direction of the net force.

Note that you could also express the acceleration as a vector if this is desired.

#veca=vecF/m#

#=1/14*<8,10>#

#=<4/7,5/7># # m//s^2#

You could then find the vector's magnitude:

#a=sqrt((a_x)^2+(a_y)^2)#

#=sqrt((4/7)^2+(5/7)^2)#

#=0.91# #m//s^2#