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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A tire weighing $2Kgs$ begins rolling down a steep hill (a $30^{\circ}$ incline) as shown. Determine the acceleration of the tire. Ignore resistance force.

$\begin{array}{1 1}4.9 \;m/s^2 \\9.8\;m/s^2\\6.4 \;m/s^2\\11.2\;m/s^2\end{array} $

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Answer : $4.9 \;m/s^2$
The $F_{grav}$ can be calculated from the mass of the object .
$F_{grav} =m.g =(2 \;kg) . (9.8 m/s/s) =19.6 \;N$
The parallel and perpendicular components of the gravity force can be determined from their respectively equations :
$F_{parallel} =19.6\;N . \sin (30\; degrees)=9.8\;N$
$F_{perpendicular}=19.6 \;N. \cos (30 \;degrees)=17.0\;N$
The acceleration of the tire is the ratio of net force to mass :
$a= F_{net}/m =(9.8 \;N) /(2\;kg) =4.9 m/s/s$
answered Aug 27, 2014 by meena.p
 

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