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16 days ago

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

f 17 newtons west. What is the magnitude of the acceleration of the box?

2 answers:

Sav [2.2K]16 days ago

8 0

The acceleration of the box has a magnitude of 2.0 m/s²

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Expanded explanation

Newton's second law of motion indicates that the total force acting on an object is directly proportional to the mass and acceleration of that object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Mass of the Object ( kg )

a = Acceleration ( m )

Now, let’s approach this problem!

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Provided:

mass of box = m = 5.0 kg

force applied = F = 27 N

frictional force = f = 17 N

Query:

find the box's acceleration = a =?

Solution:

We will apply Newton's Law of Motion to resolve this:

$\Sigma F = ma$

$F - f = ma$

$27 - 17 = 5.0a$

$10 = 5.0a$

$a = 10 \div 5.0$

$a= 2.0 \texttt{ m/s}^2$

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Final answer:

The box's acceleration has a magnitude of 2.0 m/s²

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Further learning

Effects of Gravity:
Impact of Earth’s Gravity on Objects:
Acceleration Due To Gravity:
Newton's Laws of Motion:
Illustration of Newton's Laws:

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Answer Clarifications

Grade Level: High School

Subject Focus: Physics

Chapter: Dynamics

Yuliya22 [2.4K]16 days ago

6 0

Answer:

The box experiences an acceleration with a magnitude of 2 m/s².

Explanation:

Details provided:

Mass of the box: $m=5.0$ kg

Force acting towards the east: $F=27$ N

Frictional force acting towards the west: $f=17$ N

Let’s denote the acceleration as $a$ m/s².

Thus, the net force acting on the box in the east direction is stated as:

$F_{net}=F-f=27-17=10\textrm{ N}$

According to Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Thus, the magnitude of the box's acceleration is determined to be 2 m/s².

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