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Newton's Second law of motion :- The rate of change of momentum is directly proportional to the force applied on the system.
Force applied is directly proportional to the product of mass and acceleration .
Let be the initial and final momentums respectively.
According to newton's second law :-
pf - pi / t ∝ F
We know that, Momentum ( P) = mv .
Let v be the final and u be the initial velocity .
Now,
mv - mu / t ∝ F
F ∝ m ( v-u) /t
F ∝ ma.
F = kma.
Here, K is the proportionality constant. It's value is 1 .
Units of Force are given by the units of mass and acceleration. Units of force is Kgm/s² .
In accordance to honour the contributions of Newton, 1 kgm/s² is termed as 1 Newton.
Force applied is directly proportional to the product of mass and acceleration .
Let be the initial and final momentums respectively.
According to newton's second law :-
pf - pi / t ∝ F
We know that, Momentum ( P) = mv .
Let v be the final and u be the initial velocity .
Now,
mv - mu / t ∝ F
F ∝ m ( v-u) /t
F ∝ ma.
F = kma.
Here, K is the proportionality constant. It's value is 1 .
Units of Force are given by the units of mass and acceleration. Units of force is Kgm/s² .
In accordance to honour the contributions of Newton, 1 kgm/s² is termed as 1 Newton.
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The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Derivation of Newton's second law of motion
Suppose an object of mass, m is moving along a straight line with an initial velocity, u.
It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F
throughout the time, t. The initial and final momentum of the object will be,
p1 = mu and p2 = mv respectively.
The change in momentum ∝ p2 – p1
The change in momentum∝ mv – mu
The change in momentum∝ m × (v – u).
The rate of change of momentum ∝ m × (v −u)/t
Or, the applied force,
F ∝m × (v −u)/t
F = km (v - u)/t
F = kma
Here a [a = (v – u)/t ] is the acceleration, which is the rate of change of velocity. The quantity, k is a constant of proportionality.The SI units of mass and acceleration are kg and m s-2respectively. The unit of force is so chosen that the value of the constant, k becomes one. For this, one unit of force is
defined as the amount that produces an acceleration of 1 m s-2 in an object of 1 kg mass.
That is, 1 unit of force = k × (1 kg) × (1 m s-2).
Thus, the value of k becomes 1. and
F = ma which is the mathematical expression on the Newton's second law of motion
Derivation of Newton's second law of motion
Suppose an object of mass, m is moving along a straight line with an initial velocity, u.
It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F
throughout the time, t. The initial and final momentum of the object will be,
p1 = mu and p2 = mv respectively.
The change in momentum ∝ p2 – p1
The change in momentum∝ mv – mu
The change in momentum∝ m × (v – u).
The rate of change of momentum ∝ m × (v −u)/t
Or, the applied force,
F ∝m × (v −u)/t
F = km (v - u)/t
F = kma
Here a [a = (v – u)/t ] is the acceleration, which is the rate of change of velocity. The quantity, k is a constant of proportionality.The SI units of mass and acceleration are kg and m s-2respectively. The unit of force is so chosen that the value of the constant, k becomes one. For this, one unit of force is
defined as the amount that produces an acceleration of 1 m s-2 in an object of 1 kg mass.
That is, 1 unit of force = k × (1 kg) × (1 m s-2).
Thus, the value of k becomes 1. and
F = ma which is the mathematical expression on the Newton's second law of motion
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