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= <5 N , 7 N># and the second is #F_2 = < 8 N, -2 N>#. What is the object's rate and direction of acceleration?

Emily Abbate · Accepted Answer

#a=6.97" "m/s^2#

#F_1=<5,7>" ; "F_2=<8,-2>#

#F_R:"Resultant force"#

#F_R=(5+8)i+(7-2)j#

#F_R=13i+5j#

#"The magnitude of "F_R " can be calculated using:" #

#F_R=sqrt(13^2+5^2)#

#F_R=sqrt(169+25)#

#F_R=sqrt194#

#F_R=13.93" N"#

#a:"acceleration m:mass of object"#

#a=F_R/m#

#a=(13.93)/2#

#a=6.97" "m/s^2#

Charles Aeschliman · Answer

undefined The net force acting on the object is $ F_{net} = F_1 + F_2 = <5 + 8, 7 - 2> = <13, 5> \, \text{N} $. Using Newton's second law, $ F_{net} = m \cdot a $, where $ m = 2 \, \text{kg} $, we can find the acceleration: $ a = \frac{F_{net}}{m} = \frac{<13, 5>}{2} = <6.5, 2.5> \, \text{m/s}^2 $. The rate of acceleration is $ \sqrt{6.5^2 + 2.5^2} \approx 6.92 \, \text{m/s}^2 $. The direction of acceleration is given by the angle $ \theta $, where $ \tan \theta = \frac{2.5}{6.5} $, so $ \theta \approx 21.8^\circ $ above the positive x-axis.

An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= &lt;5 N , 7 N&gt;# and the second is #F_2 = &lt; 8 N, -2 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= <5 N , 7 N># and the second is #F_2 = < 8 N, -2 N>#. What is the object's rate and direction of acceleration?