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Sindrei [870]
4 years ago
11

An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?

Physics
2 answers:
Nikitich [7]4 years ago
4 0

Answer:

The acceleration of the object is 4 m/s²

Explanation:

Newton's Second Law, also called the Fundamental Law of Dynamics, indicates that: "The change of motion is directly proportional to the printed driving force and occurs along a straight line along which that force is printed." This means that the acceleration of a moving object depends on the amount of force applied at a given time, in order to modify its trajectory or speed.

In summary, this law establishes that the net force applied to a body is proportional to the acceleration that body acquires. The constant of proportionality is the mass of the body, and this law can be expressed as:

F = m * a

where:

  • F=Force
  • m=mass
  • a=acceleration

In this case it is necessary to know the acceleration, then:

a=\frac{F}{m}

An object with a mass of 20 kg has a net force of 80 N acting on it. So,

  • F= 80 N
  • m= 20 kg

Replacing:

a=\frac{80 N}{20 kg}

Knowing that N is a unit of measure that is obtained as kg * m/s²:

a= 4 m/s²

<u><em>The acceleration of the object is 4 m/s²</em></u>

inysia [295]4 years ago
3 0

Answer:

2 ) 4 ms^{-2}

Explanation:

By Applying Newton's second law of motion

Force = mass * acceleration

acceleration=\frac{force}{mass} \\=\frac{80}{20} \\= 4 ms^{-2}

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
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Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

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