A 10.0-kg object is initially moving with a velocity of 20.0 m/s to the north and is acted on by a constant net force. after the object moves 30.0 m to the north, its velocity is 12.0 m/s north. what is the constant net force acting on the object?

The direction of motion of the object did not change, so we can find the force that acted on the object by using the work-energy theorem, which states that the work done by the force is equal to the variation of kinetic energy of the object:
[tex]W=K_f - K_i[/tex]
[tex]Fd= \frac{1}{2}mv_f^2- \frac{1}{2}mv_i^2 [/tex]
where
F is the force
d is the distance through which the force has been applied (30.0 m)
[tex]m=10.0 kg[/tex] is the object mass
[tex]v_i=20.0 m/s[/tex] is the initial velocity of the object
[tex]v_f=12.0 m/s[/tex] is its final velocity

Re-arranging the formula, we find the magnitude of the force:
[tex]F= \frac{m(v_f^2-v_i^2)}{2d}= \frac{(10.0 kg)((12m/s)^2-(20m/s)^2)}{2 \cdot 30.0 m}=-42.7 N [/tex]
And the negative sign means the force is in the opposite direction of the motion of the object (in fact, the object is decelerating)

johanrusli johanrusli · Answer 2 · 2019-12-20T09:16:02+01:00

The constant net force acting on the object is about 42.7 N to the south.

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of object = m = 10.0 kg

initial velocity = u = 20.0 m/s

final velocity = v = 12.0 m/s

distance = d = 30.0 m

Asked:

acceleration of the box = a = ?

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = ma[/tex]

[tex]\Sigma F = m( v^2 - u^ ) \div (2d)[/tex]

[tex]\Sigma F = 10.0( 12.0^2 - 20.0^2 ) \div (2(30.0))[/tex]

[tex]\Sigma F = 10.0( -256) \div (60.0)[/tex]

[tex]\Sigma F = -2560) \div (60.0)[/tex]

[tex]\Sigma F = -42\frac{2}{3} \texttt{ N}[/tex]

[tex]\Sigma F \approx -42.7 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

Learn more

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The Acceleration Due To Gravity : https://brainly.com/question/4189441
Newton's Law of Motion: https://brainly.com/question/10431582
Example of Newton's Law: https://brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics