Respuesta :
Answer:
(c) 0.77 m/s^2 directed 35° south of west.
Explanation:
Let's first find the resultant force and its direction:
[tex]F = \sqrt{(F_S)^2 + (F_W)^2}[/tex]
[tex]F = \sqrt{(12)^2+(17)^2}[/tex]
F = 20.808 N
To find the direction, we can just imagine the forces as lengths of a right angled triangle.
So, Force (west) will be our perpendicular.
and Force (south) will be our base.
Calculating the angle we have:
[tex]Tan(Theta) = \frac{17}{12}[/tex]
Theta = 54.78° (angle from south)
Direction of resultant force = 90 - 54.78 = 35.22° (south of west)
Taking out the acceleration, we get:
F = m * a
20.808 = 27 * a
a = 0.77 m/s^2
So the answer is (c)
The required magnitude of acceleration is [tex]0.77 \;\rm m/s^{2}[/tex] at 55 degrees from south of west. Hence, option (c) is correct.
Given data:
The mass of object is, m = 27 kg.
The magnitude of force towards the south is, F = 12 N.
The magnitude of force towards the west is, F' = 17 N.
Here we will follow the Newton's second law that applied force will cause to accelerate the object along the same direction.
Then net force is,
[tex]F_{net}= \sqrt{F^{2}+F'^{2}}\\\\F_{net}= \sqrt{12^{2}+17^{2}}\\\\F_{net} = 20.808 \;\rm N[/tex]
Now, the direction will be,
[tex]tan \theta =\dfrac{F'}{F} \\\\tan \theta =\dfrac{17}{12} \\\\\theta \approx 55^{\circ}[/tex]
This is the direction of 55 degrees towards south. So, acceleration will be towards south of west.
Then,
[tex]F_{net} = m a\\\\20.808 = 27 \times a\\\\a = 0.77 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the required magnitude of acceleration is [tex]0.77 \;\rm m/s^{2}[/tex] at 55 degrees from south of west. Hence, option (c) is correct.
Learn more about the Newton's second law here:
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