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Standard XII

Physics

Acceleration in 2D

A block of ma...

Question

A block of mass $m = 1$ kg, moving on a horizontal surface with speed $v = 2 m s^{- 1}$ enters a rough patch ranging from $x = 0.10 m$ to $x = 2.01 m$ . The retarding force F on the block in this range is inversely proportional to $x$ over this range.

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Solution

Formula,

$F = m a = m v \frac{d v}{d x} = - \frac{k}{x} = - \frac{0.5}{x}$

$v d v = - 0.5 \frac{d x}{x}$

$\int_{2}^{v} v d v = 0.5 \int_{0.1}^{2.01} \frac{1}{x} d x$

$\frac{v^{2}}{2} - \frac{2^{2}}{2} = 0.5 (- ln \frac{2.01}{0.1}) = - 1.5$

$\Rightarrow v^{2} = 2 (2 - 1.5) = 1$

$∴ v = 1 m s^{- 1}$

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A block of mass m=1 kg, moving on a horizontal surface with speed v=2 ms−1 enters a rough patch ranging from x=0.10 m to x=2.01 m. The retarding force F on the block in this range is inversely proportional to x over this range.

Formula,F=ma=mvdvdx=−kx=−0.5xvdv=−0.5dxx∫v2vdv=0.5∫2.010.11xdxv22−222=0.5(−ln2.010.1)=−1.5⇒v2=2(2−1.5)=1∴v=1ms−1

A block of mass $m = 1$ kg, moving on a horizontal surface with speed $v = 2 m s^{- 1}$ enters a rough patch ranging from $x = 0.10 m$ to $x = 2.01 m$ . The retarding force F on the block in this range is inversely proportional to $x$ over this range.