Question

Derive a mathematical expression for force.........PLS ANSWER FAST I WILL MARK AS BRAINLIEST

Answer 1

$\Large\blue{\underline{\underline{\mathtt{Derivation\:of\:Force}}}}$

$\textit{Consider a body having an initial momentum}$$\mathtt{\overrightarrow{p_{1}}}$$\textit{Let it's momentum change to}$$\mathtt{\overrightarrow{p_{1}}}$$\textit{when a net force}$$\mathtt{\overrightarrow{F}}$$\textit{acts on it's during a time interval of}$$\mathtt{\Delta t}$

$\mathtt{Change\:in\:Momentum = \mathtt{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}}$

$\mathrm{\therefore Change\:in\:Momentum\:in\:unit\:time = \dfrac{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}{\Delta t}}$

$\mathrm{i.e.,}$

$\mathrm{Rate\:of\:change\:of\:Momentum= \dfrac{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}{\Delta t}}$

$\mathtt{but\:\overrightarrow{p_{2}} = m\overrightarrow{v} (\overrightarrow{v}\:is\:the\:final\:velocity)}$

$and$

$\mathtt{but\:\overrightarrow{p_{1}} = m\overrightarrow{u} (\overrightarrow{u}\:is\:the\:initial\:velocity)}$

$\mathrm{Rate\:of\:change\:of\:Momentum = \dfrac{\mathtt{m\overrightarrow{v} - m\overrightarrow{u}}}{\Delta t}}$

$\mathrm{\rightarrow Rate\:of\:change\:of\:Momentum = m\bigg(\dfrac{\mathtt{\overrightarrow{v} - \overrightarrow{u}}}{\Delta t}\bigg)}$

$\mathrm{\rightarrow Rate\:of\:change\:of\:Momentum = \dfrac{\mathtt{m\Delta\overrightarrow{v}}}{\Delta t}}$

$\textit{Here,}$$\Delta\overrightarrow{v}$$\textit{is the change of velocity}$.

$\textsf{From Newton's second law of motion,}$

$\mathtt{F \propto \dfrac{m\Delta\overrightarrow{v}}{\Delta t}}$

$\mathtt{\Rightarrow F \propto ma}$

$\mathtt{\because \overrightarrow{a} = \dfrac{\Delta v}{\Delta t}}$

$\mathtt{\Rightarrow F = k\:ma}$

$\textit{Unit of force is chosen in such a way}$ $\textit{that it produces unit acceleration in a body of unit mass.}$

$\textit{Then , the constant of proportionality, k = 1}$

$\green{\boxed{F = ma}}$

______________________________________

Answer 2

Answer:

Consider a body having an initial momentum \mathtt{\overrightarrow{p_{1}}}

p

1

\textit{Let it's momentum change to}Let it’s momentum change to \mathtt{\overrightarrow{p_{1}}}

p

1

\textit{when a net force}when a net force \mathtt{\overrightarrow{F}}

F

\textit{acts on it's during a time interval of}acts on it’s during a time interval of \mathtt{\Delta t}Δt

\mathtt{Change\:in\:Momentum = \mathtt{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}}ChangeinMomentum=

p

2

−

p

1

\mathrm{\therefore Change\:in\:Momentum\:in\:unit\:time = \dfrac{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}{\Delta t}}∴ChangeinMomentuminunittime=

Δt

p

2

−

p

1

\mathrm{i.e.,}i.e.,

\mathrm{Rate\:of\:change\:of\:Momentum= \dfrac{\overrightarrow{p_{2}} - \overrightarrow{p_{1}}}{\Delta t}}RateofchangeofMomentum=

Δt

p

2

−

p

1

\mathtt{but\:\overrightarrow{p_{2}} = m\overrightarrow{v} (\overrightarrow{v}\:is\:the\:final\:velocity)}but

p

2

=m

v

(

v

isthefinalvelocity)

andand

\mathtt{but\:\overrightarrow{p_{1}} = m\overrightarrow{u} (\overrightarrow{u}\:is\:the\:initial\:velocity)}but

p

1

=m

u

(

u

istheinitialvelocity)

\mathrm{Rate\:of\:change\:of\:Momentum = \dfrac{\mathtt{m\overrightarrow{v} - m\overrightarrow{u}}}{\Delta t}}RateofchangeofMomentum=

Δt

m

v

−m

u

\mathrm{\rightarrow Rate\:of\:change\:of\:Momentum = m\bigg(\dfrac{\mathtt{\overrightarrow{v} - \overrightarrow{u}}}{\Delta t}\bigg)}→RateofchangeofMomentum=m(

Δt

v

−

u

)

\mathrm{\rightarrow Rate\:of\:change\:of\:Momentum = \dfrac{\mathtt{m\Delta\overrightarrow{v}}}{\Delta t}}→RateofchangeofMomentum=

Δt

mΔ

v

\textit{Here,}Here, \Delta\overrightarrow{v}Δ

v

\textit{is the change of velocity}is the change of velocity .

\textsf{From Newton's second law of motion,}From Newton’s second law of motion,

\mathtt{F \propto \dfrac{m\Delta\overrightarrow{v}}{\Delta t}}F∝

Δt

mΔ

v

\mathtt{\Rightarrow F \propto ma}⇒F∝ma

\mathtt{\because \overrightarrow{a} = \dfrac{\Delta v}{\Delta t}}∵

a

=

Δt

Δv

\mathtt{\Rightarrow F = k\:ma}⇒F=kma

\textit{Unit of force is chosen in such a way}Unit of force is chosen in such a way \textit{that it produces unit acceleration in a body of unit mass.}that it produces unit acceleration in a body of unit mass.

\textit{Then , the constant of proportionality, k = 1}Then , the constant of proportionality, k = 1

\green{\boxed{F = ma}}

F=ma

gajb answer nikhil bhai