Relation between Newton and Dyne

What is Newton?

Newton (N) is the SI unit of force. It is named after the famous scientist Sir Isaac Newton, who formulated the laws of motion.

One Newton is defined as the amount of force required to accelerate a mass of one kilogram by one metre per second squared.
1 Newton = Force needed to move 1 kg of mass with an acceleration of $1 \mathrm{~m} / \mathrm{s}^2$.

$
1 N=1 \mathrm{~kg} \times 1 \mathrm{~m} / \mathrm{s}^2
$

Example:
If you push a 1 kg object and it starts moving with an acceleration of $1 \mathrm{~m} / \mathrm{s}^2$, then you are applying a force of 1 Newton.

What is Dyne?

Dyne is a unit of force in the centimetre-gram-second (CGS) system of units.
It is defined as the force required to accelerate a mass of one gram by one centimetre per second squared.

1 Dyne $=$ Force needed to move 1 gram of mass with an acceleration of $1 \mathrm{~cm} / \mathrm{s}^2$

$
1 \text { dyne }=1 g \times 1 c m / s^2
$
Relation with Newton:

$
1 \mathrm{~N}=10^5 \text { dyne }
$

Example:
If you apply a small force that makes a 1 g object move faster by 1 cm every second, you are applying a force of 1 dyne.

Relation Between Newton and Dyne

The relation between Newton and Dyne is based on their definitions in the SI and CGS systems of units.
1 Newton ( $\mathbf{N}$ ) = Force required to accelerate $\mathbf{1} \mathbf{~ k g}$ mass by $\mathbf{1 ~ m} / \mathbf{s}^{\mathbf{2}}$.
$\mathbf{1}$ Dyne $=$ Force required to accelerate $\mathbf{1} \mathbf{~ g}$ mass by $\mathbf{1} \mathbf{~ c m} / \mathbf{s}^{\mathbf{2}}$.

Now,

$
\begin{aligned}
& 1 \mathrm{~kg}=1000 \mathrm{~g} \\
& 1 \mathrm{~m}=100 \mathrm{~cm}
\end{aligned}
$

So,

$
\begin{gathered}
1 N=1000 g \times\left(100 c m / s^2\right) \\
1 N=10^5 d y n e
\end{gathered}
$

Also, read

• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for All Subjects

Difference Between Newton and Dyne

Applications of Converting Newton to Dyne

• It helps in converting force between SI system (Newton) and CGS system (dyne) for uniform calculations.
• It is widely used in numerical problems of physics, especially in chapters like mechanics and units & dimensions.
• It is useful in scientific research and laboratory experiments, where small forces are better expressed in dynes.
• It helps engineers in designing machines and electrical devices, where accurate force calculation is required.
• It improves the understanding of fundamental concepts like force, mass, and acceleration in different unit systems.

Solved Examples: Converting Newton to Dyne

Example 1: A constant force of $\mathbf{3} \mathbf{N}$ is applied on a body. Calculate the value of this force in dyne.
Solution:
We know that:

$
\begin{gathered}
1 N=10^5 \text { dyne } \\
\text { Force }=3 \times 10^5 \text { dyne }
\end{gathered}
$

Answer: $3 \times 10^5$ dyne

Example 2: A small object is pulled with a force of $\mathbf{0 . 2} \mathbf{N}$. Convert this force into dyne.
Solution:

$
\begin{gathered}
\text { Force }=0.2 \times 10^5 \\
=2 \times 10^4 \text { dyne }
\end{gathered}
$

Answer: $2 \times 10^4$ dyne

Example 3: A force of $\mathbf{1 5} \mathbf{~ N}$ acts on a particle moving in a straight line. Express this force in dyne.
Solution:

$
\begin{aligned}
& \text { Force }=15 \times 10^5 \\
& =1.5 \times 10^6 \text { dyne }
\end{aligned}
$

Answer: $1.5 \times 10^6$ dyne

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