An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= # and the second is #F_2 = #. What is the object's rate and direction of acceleration?

Question

An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= < -9 N , 8 N># and the second is #F_2 = < 4 N, -1 N>#. What is the object's rate and direction of acceleration?

Nora Arzate · Accepted Answer

The rate of acceleration is #=4.3ms^-2# in the direction of #=125.5#º

The resultant force is #vecF=vecF_1+vecF_2# #= <-9,8>+<4,-1># #=<-5,7># We apply Newton's second Law #vecF=m veca# #veca =1/m*vecF# #=1/2<-5,7> = <-5/2, 7/2># The magnitude of the acceleration is #||veca||=||<-5/2,7/2>||# #=sqrt((-5/2)^2+(7/2)^2)# #=sqrt(74)/2=4.3ms^-2# The direction is #theta = arctan((7)/(-5))# The angle is in the 2nd quadrant #theta=arctan(-7/5)=125.5 #º

Adam Arnold · Answer

undefined The object's acceleration is $ a = \frac{F_{net}}{m} $, where $ F_{net} $ is the net force acting on the object and $ m $ is the mass of the object. The net force can be found by adding the individual forces $ F_1 $ and $ F_2 $ together. So, $ F_{net} = F_1 + F_2 $.

$ F_{net} = (-9 \, N, 8 \, N) + (4 \, N, -1 \, N) = (-5 \, N, 7 \, N) $

Now, $ F_{net} = (-5 \, N, 7 \, N) $ and $ m = 2 \, kg $, so

$ a = \frac{F_{net}}{m} = \frac{(-5 \, N, 7 \, N)}{2 \, kg} = (-2.5 \, m/s^2, 3.5 \, m/s^2) $.

Therefore, the object's rate of acceleration is $ 2.5 \, m/s^2 $ in the negative x-direction and $ 3.5 \, m/s^2 $ in the positive y-direction.

An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= &lt; -9 N , 8 N&gt;# and the second is #F_2 = &lt; 4 N, -1 N&gt;#. What is the object's rate and direction of acceleration?

An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= < -9 N , 8 N># and the second is #F_2 = < 4 N, -1 N>#. What is the object's rate and direction of acceleration?