An object with a mass of #14 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 #14 kg# is acted on by two forces. The first is #F_1= < 2 N , 7 N># and the second is #F_2 = < -9 N, 4 N>#. What is the object's rate and direction of acceleration?

Nora Arzate · Accepted Answer

The rate of acceleration is #=0.93ms^-2# in the direction #57.5#º

The resultant force is #vecF=vecF_1+vecF_2# #= <2,7>+<-9,4># #=<-7,11># we use Newton's second Law #vecF=m veca# #veca =1/m*vecF# #=1/14<-7,11> = <-1/2, 11/14># The magnitude of the acceleration is #||veca||=||<-1/2,11/14>||# #=sqrt((1/2)^2+(11/14)^2)# #=sqrt(1/4+121/196)# #=sqrt 0.867=0.93ms^-2# The direction is #theta = arctan((11/14)/(1/2))# The angle is in the 1st quadrant #theta=arctan(11/7)=57.5#º

Axel Acevedo · Answer

undefined The net force acting on the object can be found by adding the two given forces together: F_net = F_1 + F_2 = <2 N - 9 N, 7 N + 4 N> = <-7 N, 11 N>. Using Newton's second law, F_net = m * a, where m is the mass of the object and a is the acceleration. Thus, the acceleration can be calculated as: a = F_net / m = <-7 N / 14 kg, 11 N / 14 kg> = <-0.5 m/s^2, 0.79 m/s^2>. So, the rate of acceleration is -0.5 m/s^2 in the x-direction and 0.79 m/s^2 in the y-direction.

An object with a mass of #14 kg# is acted on by two forces. The first is #F_1= &lt; 2 N , 7 N&gt;# and the second is #F_2 = &lt; -9 N, 4 N&gt;#. What is the object's rate and direction of acceleration?

An object with a mass of #14 kg# is acted on by two forces. The first is #F_1= < 2 N , 7 N># and the second is #F_2 = < -9 N, 4 N>#. What is the object's rate and direction of acceleration?