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?

Answer 1

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
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Answer 2
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.
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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.

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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