### Newton’s second law and equation of Force

##### September 24, 2018

# Newton’s second law

Newton’s second law states that the rate of change of momentum of a body is equal to the net force applied on it. So it’s evident that the second law of Newton was stated in terms of momentum.Mathematically it can be stated this way: **F _{net }= (mV-mU)/t**. Here, U is the initial velocity of the mass m. F

_{net}is the net force being applied on that object for a time duration t. After time t the velocity of that object becomes V.

**Newton’s 3rd Law**

## Newton’s second law – equation of force

Initial momentum = mU and final momentum = mV.

So change of momentum = mV – mU

And rate of change of momentum = **(mV – mU) / t**

So as per*Newton’s 2nd law*,

F_{net }= (mV – mU) / t

or, F_{net}= m (V – U) / t = m a [acceleration a = rate of change of velocity = (V – U) / t ]

so, **F _{net} = m a**

**Net Force = mass X acceleration**

Thus we get the **equation of Force** from **Newton’s 2 ^{nd} Law** of motion. This helps us in measurement of force.

## Newton’s 2nd law – a few words

1. Net force means **unbalanced force**. There may be multiple force components working simultaneously on a body. Few of them may cancel each other. After getting the vector summation of all force components if we get a non-zero resultant force then we can say that a net force is working on the body.

2. Net force means acceleration of the body. Acceleration means change in velocity with time.

3. If force is zero, then from **F =ma** we get **ma = 0**

As m is non-zero, so acceleration a = 0 .

**So in absence of force there is no acceleration.**

Again when acceleration is zero that means no change in velocity. That means the object in motion moves with **uniform velocity**. We will solve few numerical problems now, in next paragraph.

## Newton’s 2nd law – sample problems

1) **Force of 1000 Newton is applied on a 25 kg mass for 4 seconds. What would be its velocity?**

**Solution:
**

F =1000 N

m=25kg

t= 4 sec

u=0

v=?

Acceleration = a = F/m = 1000/25 = 40 m/s^{2 }

v = u + at =0 + 40X4 = 160 m/s

2) **A force is applied on a mass of 20 kg for 3 seconds. As the force is removed the mass moves 81 meter in 3 seconds. What was the value of the force applied?**

Solution:

In first 3 secs when the Force is present the body moves with acceleration and attains a velocity. After these first 3 seconds the force is removed. And in the next 3 secs when Force is not there, the body moves with a uniform velocity.

From the data we can find the uniform velocity for the next 3 secs:

V = 81/3 = 27 m/s

This velocity was attained in the first 3 seconds when the force was present and the mass was moving with acceleration.

So we can calculate the acceleration in first 3 seconds. **a** = (27-0)/3 = 9 m/s².

Therefore the force applied on the mass for first 3 secs= mass x acceleration = 20 x 9 = 180 N

**Contact forces between 2 blocks: Analysis using Newton’s Laws**

To read the article in this Blog, pls follow the link: **Contact forces between 2 blocks: Analysis using Newton’s Laws**

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