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4-1.Newton's Laws of Motion
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A block is kept on a fixed smooth wedge whose vertical section is a curve $y = \frac{{{x^2}}}{{\sqrt 3 }}$ as shown in figure where $x$ represents horizontal direction and $y$ represents vertical direction. When released from a point where $y = \frac{1}{{4\sqrt 3 }}$ , what will be its acceleration ? $(g \,= 10\, m/s^2)$ ....... $m/s^2$
A$2.5$
B$5\sqrt 3 $
C$5$
DCan’t be determined
Solution
$\frac{d y}{d x}=\tan \theta=\frac{2 x}{\sqrt{3}} \Rightarrow \tan \theta=\frac{2}{\sqrt{3}} \cdot \frac{1}{2}=\frac{1}{\sqrt{3}}$
acceleration along wedge
$=g \sin \theta=10 \cdot \frac{1}{2}=5 \mathrm{m} / \mathrm{s}^{2}$
acceleration along wedge
$=g \sin \theta=10 \cdot \frac{1}{2}=5 \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
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