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, -3 N >>#. What is the objects rate and direction of acceleration?

Answer 1

The rate of acceleration is #=0.4ms^-1# in the direction of #=213.7º# anticlockwise from the x-axis

The resultant force is

#vecF=vecF_1+vecF_2#
#= <-7,1>+<4,-3>#
#=<-3,-2>#

We apply Newton's second Law

#vecF=m veca#
Mass, #m=9kg#
#veca =1/m*vecF#
#=1/9<-3,-2> = <-1/3, -2/9>#

The magnitude of the acceleration is

#||veca||=||<-1/3,-2/9>||#
#=sqrt((-1/3)^2+(-2/9)^2)#
#=sqrt(0.16049)=0.4ms^-2#
The direction is #theta = arctan((2)/(3))#

The angle lies in the 3rd quadrant

#theta=arctan(2/3)=180+33.7º=213.7º# anticlockwise from the x-axis
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Answer 2

To find the object's acceleration, use Newton's second law: ( F_{\text{net}} = m \cdot a ). The net force is the vector sum of ( F_1 ) and ( F_2 ). Calculate ( F_{\text{net}} ), then find acceleration (( a )) using ( a = \frac{F_{\text{net}}}{m} ). The direction of acceleration is given by the vector ( a ).

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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 the given forces and then apply Newton's second law of motion, ( F = ma ), where ( F ) is the net force, ( m ) is the mass of the object, and ( a ) is the acceleration.

First, find the net force by summing the individual forces:

( F_{\text{net}} = F_1 + F_2 = (-7 , \text{N}, 1 , \text{N}) + (4 , \text{N}, -3 , \text{N}) )

( F_{\text{net}} = (-7 + 4 , \text{N}, 1 - 3 , \text{N}) )

( F_{\text{net}} = (-3 , \text{N}, -2 , \text{N}) )

Now, calculate the magnitude of the net force:

( F_{\text{net}} = \sqrt{(-3 , \text{N})^2 + (-2 , \text{N})^2} )

( F_{\text{net}} = \sqrt{9 + 4} )

( F_{\text{net}} = \sqrt{13} , \text{N} )

Next, apply Newton's second law to find the acceleration:

( F_{\text{net}} = ma )

( \sqrt{13} , \text{N} = 9 , \text{kg} \times a )

( a = \frac{\sqrt{13} , \text{N}}{9 , \text{kg}} )

Now, you have the magnitude of acceleration. To find the direction, you can calculate the angle using trigonometric functions:

( \theta = \arctan{\frac{-2 , \text{N}}{-3 , \text{N}}} )

( \theta = \arctan{\frac{2}{3}} )

( \theta \approx 33.69^\circ )

So, the object's rate of acceleration is ( \frac{\sqrt{13}}{9} , \text{m/s}^2 ) at an angle of approximately ( 33.69^\circ ) below the negative x-axis.

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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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