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?

2 Answers
Mar 28, 2017

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

Mar 28, 2017

#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#