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جستجو

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The diagram shows the vertical cross-section $ABC$ of a fixed surface. $AB$ is a curve and $BC$ is a horizontal straight line. The part of the surface containing $AB$ is smooth and the part containing $BC$ is rough. $A$ is at a height of $1.8{\text{ }}m$ above $BC$. A particle of mass $0.5{\text{ }}kg$ is released from rest at $A$ and travels along the surface to $C$.

a) Find the speed of the particle at $B$.

b) Given that the particle reaches $C4$ with a speed of $5{\text{ }}m{\text{ }}{s^{ - 1}}$, find the work done against the resistance to motion as the particle moves from $B$ to $C$. 

پاسخ تشریحی :
نمایش پاسخ

a) $\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}{v^2} = 10x1.8} \right]$

Speed is $6m{s^{ - 1}}$

b) [$WD = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}x0.5\left( {{6^2} - {5^2}} \right)$ or

$0.5x10x1.8 = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}x0.5x{5^2}$]

Work done is $2.75{\text{ }}J$

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