An object with a mass of #9 kg# is acted on by two forces. The first is #F_1= < -7 N , 1 N># and the second is #F_2 = < 4 N, -5 N>#. What is the objects rate and direction of acceleration?

Answer 1

#a = 5/9# #"m/s"^2#

#theta = 233^"o"#

We're asked to find the acceleration (magnitude and direction) of an object with a known mass and two forces that act on it.

We can use the famous equation

#sumvecF = mveca#

to help solve this problem. In component form, this becomes

#sumF_x = ma_x#
#sumF_y = ma_y#

We therefore need to find the components of the net force acting on the object.

The net force components are the vector sums of the components of the individual forces:

#sumF_x = -7# #"N"# #+ 4# #"N"# #= -3# #"N"#
#sumF_y = 1# #"N"# #- 5# #"N"# #= -4# #"N"#

The components of the acceleration are

#a_x = (sumF_x)/m = (-3color(white)(l)"N")/(9color(white)(l)"kg") = -1/3# #"m/s"^2#
#a_y = (sumF_y)/m = (-4color(white)(l)"N")/(9color(white)(l)"kg") = -4/9# #"m/s"^2#

The magnitude of the acceleration is thus

#a = sqrt((a_x)^2 + (a_y)^2) = sqrt((-1/3color(white)(l)"m/s"^2)^2 + (-4/9color(white)(l)"m/s"^2)^2)#
#= color(red)(5/9# #color(red)("m/s"^2#

The direction of the acceleration is

#theta = arctan((a_y)/(a_x)) = arctan((-4/9cancel("m/s"^2))/(-1/3cancel("m/s"^2))) = 53.1^"o" + 180^"o"#
#= color(blue)(233^"o"#
(The #180^"o"# was added because the acceleration is directed toward the third quadrant relative to the object.)
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Answer 2
The net force acting on the object is F_net = F_1 + F_2 = <-3 N, -4 N>. To find the acceleration, we use Newton's second law: F_net = m * a, where F_net is the net force, m is the mass, and a is the acceleration. Thus, a = F_net / m. Substituting the values, we get a = <-3 N / 9 kg, -4 N / 9 kg> = <-0.33 m/s^2, -0.44 m/s^2>. Therefore, the object's rate of acceleration is approximately 0.33 m/s^2 in the horizontal direction and 0.44 m/s^2 in the vertical 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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