Question

A block is kept on a smooth wedge whose vertical section is a curve $y=x^2/\surd 3$ 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 accelerations ? $(g=10m/s^2)$

Answer 1

At the releasing point acceleration will be $g\sin \theta$, for calculating we can use the concept that , $\tan \theta =\frac{y}{x}=\frac{1/4\sqrt{3}}{1/2}=\frac{1}{2\sqrt{3}}$

since , if $y=1/4\sqrt{3},\Rightarrow x=1/2$

therefore , $\sin \theta =p/h=1/\sqrt{13}$

we get the value of acceleration as , $a=g\times \frac{1}{\sqrt{13}}=2.78m/s^2$