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Q. A body of mass $2 \,kg$ slides down a curved track which is quadrant of a circle of radius $1\,m$ (figure) All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track isPhysics Question Image

Chhattisgarh PMTChhattisgarh PMT 2010

Solution:

By applying law of conservation of energy $ mgR=\frac{1}{2}mv^{2} $
$ \Rightarrow $ $ v=\sqrt{2Rg} $
$ =\sqrt{2\times 1\times 10} $
$ =4.43\,m/s $