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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 _{$y = \frac{x^{3}}{30} - \frac{x^{2}}{10}$} . 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

$y = \frac{x^{3}}{30} - \frac{x^{2}}{10}$

$\frac{d y}{d x} = \frac{x^{2}}{10} - \frac{x}{5} a n d a t x = 2; \frac{d y}{d x} = 0$

$\frac{d^{2} y}{d x^{2}} = \frac{x}{5} = \frac{1}{5} a n d a t x = 2; \frac{d^{2} y}{d x^{2}} = + v e$

Hence the particle is at it's lowest (minimum) position, in it's path at x = 2 m

$\frac{d^{2} y}{d t^{2}} = 2 x \frac{d x}{d t} - 2 \frac{d x}{d t}$

$(\frac{d^{2} y}{d t^{2}}) = 20 m / s^{2}$

$\Rightarrow N - m g = m \frac{d^{2} y}{d t^{2}} \Rightarrow N = 1 (10 + 20) = 30 N$

$P = f . v = (μ N) V = \frac{1}{100} \times 30 \times 10 = 3 w a t t$

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