A block of mass 1 kg is at rest on a horizontal surface at x = 0 . A force of 6 N is applied on block toward positive x-axis. Coefficient of friction between block and surface is variable and depends

Question

A block of mass 1 kg is at rest on a horizontal surface at $x = 0$ . A force of 6 N is applied on block toward positive x-axis. Coefficient of friction between block and surface is variable and depends on the its distance x from origin as $μ = (\frac{2 - x}{10})$ . If ' v ' is speed of block at $x = 2$ , find $\frac{v ^{2}}{5}$ .

Take g = 10 m/s².

TestHub · Accepted Answer

Using g = 10 m/s², the friction force varies with position.

$f (x) = μ (x) m g = 2 - x N$

The net forward force is

$F_{net} = 6 - (2 - x) = 4 + x$

Apply work-energy theorem from x = 0 to x = 2 m.

$\frac{1}{2} (1) v^{2} = \int_{0}^{2} (4 + x) d x = 10 J$

$v^{2} = 20 m^{2} / s^{2}$

$\frac{v ^{2}}{5} = 4$

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