If F = 4.0 N and m = 2.0 kg, what is the magnitude a of the acceleration for the block shown below? The surface is frictionless. *

The magnitude of the acceleration of the block is 3.5 [tex]\bold{m/s^2.}[/tex]
The correct option is (C) 3.5 [tex]\bold{m/s^2.}[/tex]
Acceleration is the rate of change of velocity with time.
By Newton’s Second law
This law establishes the relationship between the net applied force on an object with the resultant acceleration. For a constant magnitude of force, the more is the mass of the object, the lesser will be its acceleration.
Given,
The mass of the block is 2.0 kg.
The force in the horizontal direction on the block is [tex]\bold{4.0 N + 4.0 N \cos 40{}^\circ.}[/tex]
a is the acceleration in the horizontal direction.
The acceleration for the block can be determined using Newton’s Second law,
[tex]\begin{aligned} \Sigma F&=ma \\ F+F\cos 40{}^\circ &=ma \\ a&=\frac{F\left( 1+\cos 40{}^\circ \right)}{m} \\ &=\frac{4.0\text{ N}\left( 1.766 \right)}{2.0\text{ kg}} \\ &=3.5\text{ m/}{{\text{s}}^{2}} \end{aligned}[/tex]
Thus, the acceleration is 3.5 [tex]\bold{m/s^2.}[/tex]
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