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PolarNik
2 months ago
14

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?

Physics
2 answers:
Sav [3.1K]2 months ago
8 0

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

\texttt{ }

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!

\texttt{ }

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

\texttt{ }

Final answer:

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

\texttt{ }

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:

\texttt{ }

Answer Clarifications

Grade Level: High School

Subject Focus: Physics

Chapter: Dynamics

Yuliya22 [3.3K]2 months 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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