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An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= < -2 N , 8 N># and the second is #F_2 = < 4 N, -1 N>#. What is the object's rate and direction of acceleration?

Answer 1

Amelia Adams

The rate of acceleration is #=3.57ms^-2# in the direction of #=75.1#º

The resultant force is

#vecF=vecF_1+vecF_2#

#= <-2,8>+<4,-1>#

#=<2,7>#

We apply Newton's second Law

#vecF=m veca#

Mass, #m=2kg#

#veca =1/m*vecF#

#=1/2<2,7> = <1, 7/2>#

The magnitude of the acceleration is

#||veca||=||<1,7/2>||#

#=sqrt((1)^2+(7/2)^2)#

#=sqrt(51)/2=3.57ms^-2#

The direction is #theta = arctan((7)/(2))#

The angle is in the 1st quadrant

#theta=arctan(7/2)=75.1#º

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Answer 2

Adrian Ayala

To find the object's acceleration, we first need to calculate the net force acting on it by adding the individual forces. F_net = F_1 + F_2 = <-2 N, 8 N> + <4 N, -1 N> = <(-2 + 4) N, (8 - 1) N> = <2 N, 7 N> Next, we use Newton's second law, which states that F_net = m * a, where m is the mass of the object and a is its acceleration. Therefore, acceleration (a) = F_net / m = <2 N / 2 kg, 7 N / 2 kg> = <1 m/s², 3.5 m/s²> So, the object's rate of acceleration is 1 m/s² in the x-direction and 3.5 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.

An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= &lt; -2 N , 8 N&gt;# and the second is #F_2 = &lt; 4 N, -1 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= < -2 N , 8 N># and the second is #F_2 = < 4 N, -1 N>#. What is the object's rate and direction of acceleration?