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جستجو

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A particle of mass $0.8{\text{ }}kg$ slides down a rough inclined plane along a line of greatest slope $AB$. The distance $AB$ is $8{\text{ }}m$. The particle starts at $A$ with speed $3{\text{ }}m{\text{ }}{s^{ - 1}}$ and moves with constant acceleration $2.5{\text{ }}m{\text{ }}{s^{ - 2}}$.

a) Find the speed of the particle at the instant it reaches $B$.

b) Given that the work done against the frictional force as the particle moves from $A$ to $B$ is $7{\text{ }}J$, find the angle of inclination of the plane.

When the particle is at the point $X$ its speed is the same as the average speed for the motion from $A$ to $B$.

c) Find the work done by the frictional force for the particle’s motion from $A$ to $X$.

پاسخ تشریحی :
نمایش پاسخ

a) $\left[ {{v^2} = {3^2} + 2 \times 2.5 \times 8} \right]$

Speed is $7{\text{ }}m{s^{ - 1}}$

b) $KE$ gain $ = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}0.8\left( {{7^2} - {3^2}} \right){\text{ }}\left( { = 16} \right)$

$PE$ loss $ = 16 + 7$

$\left[ {0.8 \times 10 \times 8\sin \alpha  = 23} \right]$

Angle is ${21.1^ \circ }$ or ${0.368^ \circ }$

ALTERNATIVELY:

$F = 7/8$

$\left[ {0.8 \times 10\sin \alpha  - F = 0.8 \times 2.5} \right]$

$0.8 \times 10\sin \alpha  - 0.875 = 0.8 \times 2.5$

Angle is ${21.1^ \circ }$ or ${0.368^ \circ }$

c) ${5^2} = {3^2} + 2 \times 2.5s{\text{ }}\left( {s = 3.2} \right)$

$[WD/7 = 3.2/8$

or $WD = 0.875 \times 3.2$

or $WD = 8 \times 3.2 \times \left( {23/64} \right)$

$ - {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}0.8\left( {{5^2} - {3^2}} \right)]$

Work done is $2.8{\text{ }}J$

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