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A lorry of mass $15000{\text{ }}kg$ climbs a hill of length $500{\text{ }}m$ at a constant speed. The hill is inclined at ${2.5^ \circ }$ to the horizontal. The resistance to the lorry’s motion is constant and equal to $800{\text{ }}N$.

a) Find the work done by the lorry’s driving force.

On its return journey the lorry reaches the top of the hill with speed $20{\text{ }}m{\text{ }}{s^{ - 1}}$ and continues down the hill with a constant driving force of $2000{\text{ }}N$. The resistance to the lorry’s motion is again constant and equal to $800{\text{ }}N$.

b) Find the speed of the lorry when it reaches the bottom of the hill.

پاسخ تشریحی :
نمایش پاسخ

a) Gain in $PE = 15000g \times 500\sin {2.5^ \circ }J$

$WD$ against the resistance $ = 800 \times 500{\text{ }}J$

$\left[ {3271454 + 400000} \right]$

Work done is $3670000{\text{ }}J$ or $3670{\text{ }}kJ$

b) Work done by $DF = 2000 \times 500{\text{ }}J$

Gain in $KE = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}15000\left( {{v^2} - {{20}^2}} \right)$

${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}15000\left( {{v^2} - {{20}^2}} \right) = 3271454 - 400000 + $

$1000000$

Speed of the lorry is $30.3{\text{ }}m{s^{ - 1}}$

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