Question

Derive the mathematical relation of Newton’s second law of motion (F =

m*a).​

Answer 1

Required :-

• To derive the mathematical relation of Newton's 2nd law of motion

Derivation :-

Before looking at the derivation . First let's know what Newton's 2nd law of motion actually states .

So,

Newton's 2nd law of motion states that ;

" The rate of change of momentum is directly proportional to the unbalance force in the direction of force "

So,

This can be represented as ;

$\tt{ F \propto \dfrac{ \Delta P}{ \Delta t} }$$\longrightarrow{\text{Equation - 1 }}$

Consider this as equation 1

Here ,

' ∆ ' this is known as " Delta "

And , Delta represents ' change '

So,

We know that ;

P denotes Momentum

So,

Force is directly proportional to the rate of change in momentum

But,

Change in momentum can be written as ;

Let,

The initial momentum ( P1 ) be mu

Final momentum ( P2 ) be mv

Time = t

So,

Rate of change in momentum = Final momentum - Initial momentum / time

[ Rate of change can be defined in simple terms as any physical quantity divide by time ]

So,

$\dfrac{ mv - mu }{t }$

Now

Substitute this value in equation 1

So,

$\tt{ F \propto \dfrac{ mv - mu }{t} }$

Take " m " common on right side

So,

$\tt{ F \propto \dfrac{ m(v - u) }{ t }}$

Recall the first equation of motion

v = u + at

Transpose u , t to the left side

So,

$\tt{ a = \dfrac{v - u }{t } }$

Substitute this value in above one

Hence,

$\tt{ F \propto ma }$

We got,

F is directly proportional to mass and acceleration .

But we need to remove the proportionality symbol

In order to remove it we need to add an proportionality constant

Hence

F = kma

Here

' k ' is the proportional constant

we can ignore this " k "

Because

1 newton of force is defined as 1 kg mass by 1 meter per second square

1 N = k x 1kg x 1m/s²

Therefore ,

we can conclude that k = 1

So,

we can ignore k

Hence,

F = ma

Here,

F = Force

m = mass

a = acceleration

So,

The mathematical relation of Newton's 2nd law of motion is derived !

Answer 2

${\purple{ \large\sf \: Newton's \: 2nd \: Law \: of \: Motion}}$

According to Newton's 2nd law of motion The rate of change of momentum of a body is directly proportional to the applied force,and takes place in the direction in which the force acts. The rate of change of momentum of a body can be obtained by dividing the Change in momentum by Time taken for change. So,It can be expressed by :

$\sf \: Force \propto \dfrac{Change \: in \: momentum}{Time \: taken}$

Consider a body of Mass m having an initial velocity u. The initial momentum of this body will be mu. Suppose a force F acts on this body for time t & causes the final velocity to become v. The final momentum of this body will be mv. Now,the change in momentum of this body is mv - mu & the time taken for this change is t. So, According to Newton's First Law of Motion :

$\large \: \sf \: F \propto \dfrac{ mv \: - \: mu}{t}$

or

$\large \: \sf \: F \propto \dfrac{ m(v - u)}{t}$

But $\sf\frac{v-u}{t}$ represents change in velocity with time which is known as acceleration a. So,by writing a in place of $\sf\frac{v-u}{t}$ in the above relation,we get :

$\large \: \sf \: F \propto m \times a$

Thus, the force acting on a body is directly proportional to the product of mass of the body and acceleration produced in the body by the action of the force, and it acts in the direction of acceleration.

The relation F ∝ m × a can be turned into an equation by putting in a constant k.

Thus,

$\large \: \sf \: F = k \times m \times a \: (where \: k \: is \: constant)$

The value of constant k in SI units is 1,so the above equation becomes

$\large \: \sf \: F = m \times a$

$\boxed{\red{\large\sf \: Force = mass \times acceleration}}$

Thus, Newton's 2nd law of motion gives us a relationship between Force & Acceleration.