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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Determine the acceleration of a 12kg block sliding down a plane surface angled at $47^{\large\circ}$ to the horizontal.Assume that friction is negligible.

$\begin{array}{1 1}3.6m/s^2\\9.8m/s^2\\14.4m/s^2\\7.2m/s^2\end{array} $

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1 Answer

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Answer : $7.2m/s^2$
Draw a free body diagram :
$F_{net} = F_{gx}$
$F_{net}=F_g(\sin \alpha)$
$\quad\quad=(12kg)(9.80m/s^2)(\sin 47^{\large\circ})$
$\quad\quad=86.0072N$
Using Newton's second law :
$\quad\quad\;\;\; F_{net}=ma$
$(86.0072N)=(12kg)(a)$
$\quad\quad\;\;\;\;\;\;\;a=7.2m/s^2$
answered Aug 27, 2014 by sreemathi.v
 

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