What is an expression for the force due to gravity on a body of mass m ?
A. F=0
B. F=$\dfrac{m}{g}$
C. F=g
D. F=mg
Answer
646.2k+ views
Hint-To solve this question, we need to know the basic theory related to Newton's law of motion. As we know Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass as discussed below.
Formula used- ${\text{a = }}\dfrac{{\text{F}}}{{\text{m}}}$
Where,
F is the force act on a body
m is the mass of that object
a is acceleration.
Complete step by step solution: -
If an object placed in the field of the gravitational pull of the Earth experiences the gravitational force. Acceleration (due to gravity) is defined as the acceleration gained by an object because of the force of gravity acting on the body. It is denoted by ‘g’ and always measured in terms of ${\text{m/}}{{\text{s}}^{\text{2}}}$. Acceleration due to gravity having both direction as well as magnitude. Thus, we say it is a vector quantity.
From above discussion we conclude that The acceleration of a moving body is directly proportional to the magnitude of the net force on the body, in the same direction as the net force, and also inversely proportional to the mass.
above statement can be expressed in equation form as follows:
${\text{a = }}\dfrac{{\text{F}}}{{\text{m}}}$
put a=g
F=mg
Above equation derives from the Newton second principle of mechanics, according to which the force acting on a free body produces an acceleration:
F=ma,
where the constant m represents the body mass. In other words, the acceleration is proportional to the applied force. If you are present in a gravitational field then the acceleration produced by gravity is called g (earth surface it is about 9.8 ${\text{m/}}{{\text{s}}^{\text{2}}}$).
Here, we can write the second law of motion as follows:
F=mg.
Therefore, F=mg is an expression for the force due to gravity on a body of mass m.
Thus, option (D) is the correct answer.
Note- Always remember that this value of g does not depend on m, the mass of the object being dropped, and in fact is constant for a given altitude above a given planet. We generally approximate the value of g to be perfectly constant for systems very near to a planet’s surface.
Formula used- ${\text{a = }}\dfrac{{\text{F}}}{{\text{m}}}$
Where,
F is the force act on a body
m is the mass of that object
a is acceleration.
Complete step by step solution: -
If an object placed in the field of the gravitational pull of the Earth experiences the gravitational force. Acceleration (due to gravity) is defined as the acceleration gained by an object because of the force of gravity acting on the body. It is denoted by ‘g’ and always measured in terms of ${\text{m/}}{{\text{s}}^{\text{2}}}$. Acceleration due to gravity having both direction as well as magnitude. Thus, we say it is a vector quantity.
From above discussion we conclude that The acceleration of a moving body is directly proportional to the magnitude of the net force on the body, in the same direction as the net force, and also inversely proportional to the mass.
above statement can be expressed in equation form as follows:
${\text{a = }}\dfrac{{\text{F}}}{{\text{m}}}$
put a=g
F=mg
Above equation derives from the Newton second principle of mechanics, according to which the force acting on a free body produces an acceleration:
F=ma,
where the constant m represents the body mass. In other words, the acceleration is proportional to the applied force. If you are present in a gravitational field then the acceleration produced by gravity is called g (earth surface it is about 9.8 ${\text{m/}}{{\text{s}}^{\text{2}}}$).
Here, we can write the second law of motion as follows:
F=mg.
Therefore, F=mg is an expression for the force due to gravity on a body of mass m.
Thus, option (D) is the correct answer.
Note- Always remember that this value of g does not depend on m, the mass of the object being dropped, and in fact is constant for a given altitude above a given planet. We generally approximate the value of g to be perfectly constant for systems very near to a planet’s surface.
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