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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) |
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Answer» At the releasing point acceleration will be gsinθ, for calculating we can use the concept that, \(\tan\theta = \dfrac{y}{x}=\dfrac{1/4\sqrt{3}}{1/2}=\dfrac{1}{2\sqrt{3}}\) \(a=g\times \dfrac{1}{\sqrt{13}}=2.78 m/s^2\) |
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