Question

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Answer 1

Answer:

$\large\bold\red{Force,F=mg}$

Explanation:

Given,

A particle has mass 'm'.

Also,

The motion of that particle is defined as,

$y = ut + \frac{1}{2} g {t}^{2}$

Clearly,

This given Equation corresponds to the Equation of motion.

It's the Equation of displacement as a function of time.

And,

we know that,

On differentiating displacement Equation wrt time,

we get,

the Equation of velocity as a function of time.

Thus,

$= > \frac{d}{dt} y = \frac{d}{dt}( ut )+ \frac{d}{dt} (\frac{1}{2} g {t}^{2} ) \\ \\ = > \frac{d}{dt} y = u (\frac{d}{dt} t) + \frac{1}{2} g ( \frac{d}{dt} {t}^{2} ) \\ \\ = > \frac{dy}{dt} = u + \frac{g}{2} \times 2t \\ \\ = > \frac{dy}{dt} = u + gt$

But,

we know that,

$\frac{dy}{dt} = velocity \: (v)$

Thus,

$= > v = u + gt$

Now,

on differentiating velocity wrt time,

we get,

$\frac{dv}{dt} = \frac{du}{dt} + g( \frac{dt}{dt} ) \\ \\ = > \frac{dv}{dt} = 0 +( g \times 1) \\ \\ = > \frac{dv}{dt} = g$

But,

we know that,

$\frac{dv}{dt} = acceleration \: (a)$

Thus,

we get,

$a = g$

Now,

we know that,

Force, F = mass × acceleration

$\large\bold{=> F = mg}$