Physics Pls answer

Question

A block is kept on a smooth wedge whose vertical section is a curve y=x²/√3 as shown in figure where x represents horizontal direction and y represents vertical direction. When released from a point where y=1/4√3, what will be its accelerations ? (g=10 m/s²)

Answer 1

At the releasing point acceleration will be gsinθ, for calculating we can use the concept that,

\(\tan\theta = \dfrac{y}{x}=\dfrac{1/4\sqrt{3}}{1/2}=\dfrac{1}{2\sqrt{3}}\)
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,

\(a=g\times \dfrac{1}{\sqrt{13}}=2.78 m/s^2\)

Comments

Internet se copy kar ke answer de diya
Pla provide proper explanation