A block of mass 2 kg moving on a horizontal surface with speed of 4 ms^-1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = - kx where k = 12 Nm^-1. The speed of the block as it just crosses the rough surface will be :

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zero

1.5 ms^-1

2.0 ms^-1

2.5 ms^-1

answer is C.

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Detailed Solution

$\begin{array}{l} a = \frac{- kx}{2} = \frac{- 12 x}{2} = - 6 x \\ \frac{vdv}{dx} = - 6 x \\ \int_{4}^{v} vdv = - \int_{\frac{1}{2}}^{3 / 2} 6 xdx \\ \frac{v^{2} - 4^{2}}{2} = - \frac{6}{2} ({(\frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}] \\ v^{2} - 16 = - 6 (\frac{9}{4} - \frac{1}{4}) \\ v^{2} = 16 - 6 \times 2 = 4 \\ v = 2 m / s \end{array}$

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