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?

Answer 1

The rate of acceleration is #=6.67ms^-2# in the direction of #=103#º

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
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Answer 2

#a=sqrt(178)/2" "m/s^2 " , "alpha ~=-89.bar 9 ^o#

#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#
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Answer 3
The net force acting on the object is \( F_{net} = F_1 + F_2 = <2 + 1, -7 - 6> = <3, -13> \) N. The object's mass is \( m = 2 \) kg. Using Newton's second law (\( F_{net} = ma \)), we can find the acceleration: \( a = \frac{F_{net}}{m} = \frac{<3, -13>}{2} = <\frac{3}{2}, -\frac{13}{2}> \) m/s². So, the object's rate of acceleration is \( \frac{3}{2} \) m/s² in the x-direction and \( -\frac{13}{2} \) m/s² in the y-direction.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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