First slide

Work

Question

A particle is moving in a straight line with retardation proportional to its displacement. Its loss of K.E for any displacement x is proportional to

Moderate

A

$x^{2}$
B

$e^{x}$
C

$x$
D

$\log e^{x}$

Solution

Given retardation $α$ displacement
   $∴ \frac{d v}{d t} α x$
   $\Rightarrow \frac{d v}{d t} = K x k i s p r o p o r t i o n l i t y c o n s \tan t$
$∴ \frac{d v}{d x} \cdot \frac{d x}{d t} = K \cdot x$

$\frac{d v}{d x} \cdot V = K \cdot x$
$V d V = K \cdot x \cdot d x$
   $_{v_{1}}^{v_{2}} \int v d v = K \cdot_{a}^{x} \int x d x$
   $∴ \frac{{v_{2}}^{2}}{2} - \frac{{v_{1}}^{2}}{2} = K \cdot \frac{x^{2}}{2}$
$\frac{m {v_{2}}^{2}}{2} - \frac{m {v_{1}}^{2}}{2} = \frac{K \cdot m}{2} x^{2}$
$K . E_{2} - K . E_{1} = \frac{m K}{α} \cdot x^{2}$
$∴ l o s s o f K . E α x^{2}$

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