Question

[\tex] <font color = blue>[/tex]
the \: motion \: of \: a \: particle \: of \: mass \: x \: is \: described \: by \: y = ut + \frac{1}{2} {gt}^{2} . \: find \: the \: force \: acting \: on \: the \: particle \:}}}}}}[/tex] ​

Answer 1

GivEn:

• Mass of a particle is x.
• Motion of particle is described by y = ut + 1/2 gt²

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To find:

• Force acting on the particle.

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SoluTion:

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${\underline{\bf{\bigstar\;As\;per\:given\; Question\;:}}}$

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$\star\;\sf y = it + \dfrac{1}{2} gt^2$

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Now, Differentiate y with respect to time (t),

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$:\implies\sf \dfrac{dy}{dt} = u + gt$

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$:\implies\sf v = u + gt\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf \dfrac{dy}{dt} = v\bigg\rgroup$

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Again, Differentiate v with respect to time (t),

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$:\implies\sf \dfrac{dv}{dt} = g$

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$:\implies\sf a = g\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf \dfrac{dv}{dt} = a\bigg\rgroup$

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Then the force is,

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$:\implies\sf f = ma$

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$\bf Here \begin{cases} & \text{m = x} \\ & \text{a = g } \end{cases}$

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$:\implies{\underline{\boxed{\bf{\pink{f = xg}}}}}\;\bigstar$

$\therefore$ Force acting on the particle is, f = xg.

Answer 2

Answer:

f=xg

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