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A small block of mass 1 kg is moving with constant speed of 10 m/s on a typical path in a x - y vertical plane whose equation is y=x330−x210 . The coefficient of friction between the block and path is 0.01. Find the magnitude of power dissipated by frictional force in watt at x = 2 m

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Detailed Solution

y=x330−x210 dydx=x210−x5 and atx=2 ; dydx=0d2ydx2=x5=15 and at x=2 ; d2ydx2=+veHence the particle is at it's lowest (minimum) position, in it's path at x = 2 md2ydt2=2xdxdt−2dxdt(d2ydt2)=20 m/s2⇒ N−mg=md2ydt2⇒ N=1(10+20)=30NP=f.v=(μN)V=1100×30×10=3 watt

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A small block of mass 1 kg is moving with constant speed of 10 m/s on a typical path in a x - y vertical plane whose equation is y=x330−x210 . The coefficient of friction between the block and path is 0.01. Find the magnitude of power dissipated by frictional force in watt at x = 2 m