An object with a mass of #2 kg# is acted on by two forces. The first is #F_1= < 2 N , 5 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 #=sqrt10/2ms^-2# in the direction #=18.4^@# clockwise from the x-axis.

The resultant force is

#vecF=vecF_1+vecF_2=<2,5> + <1,-6> = <3,-1>#
The mass is #m=2kg#

According to Newton's Second Law of motion

#vecF=mveca#

The acceleration is

#veca=1/mvecF=1/2 <3,-1> = <3/2,-1/2>#

The rate of acceleration is

#||veca|| =| |<3/2,-1/2>|| = sqrt((3/2)^2+(1/2)^2)=sqrt(9/4+1/4)=sqrt10/2ms^-2#

The direction of acceleration is

#theta = arctan((-1/2)/(3/2))=arctan(-1/3)=18.4^@# clockwise from the x-axis.
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Answer 2
The net force acting on the object is <3 N, -1 N>. The acceleration of the object is <1.5 m/s², -0.5 m/s²>.
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Answer 3
To find the object's rate and direction of acceleration, you need to calculate the net force acting on the object using vector addition, and then apply Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. First, add the two forces: F_net = F_1 + F_2 = <2 N, 5 N> + <1 N, -6 N> = <2 N + 1 N, 5 N - 6 N> = <3 N, -1 N> Now, calculate the magnitude of the net force: |F_net| = √((3 N)^2 + (-1 N)^2) = √(9 N^2 + 1 N^2) = √(10 N^2) = √10 N Next, calculate the acceleration using Newton's second law: F_net = m * a a = F_net / m Given that the mass (m) is 2 kg: a = (√10 N) / 2 kg ≈ √10 / 2 m/s^2 So, the rate of acceleration is approximately √10 / 2 m/s^2. To find the direction of acceleration, you can use the components of the net force: The direction of acceleration is given by the direction of the net force vector, which is <3 N, -1 N>. This means the object is accelerating in the direction of the vector <3 N, -1 N>.
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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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