Question

A particle moves in a straight line with retardation proportional to its displacement.its loss of kinetic energy for any displacement x is proportional to

Answer 1

deceleration is directly proportional to displacement

so we can write

$a = -kx$

$v\frac{dv}{dx} = - kx$

now integrate both sides

$\frac{v^2}{2} = -\frac{kx^2}[2}$

$v^2 = - kx^2$

now as we know that kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}m(-kx^2)$

$K = -\frac{1}{2}mkx^2$

so kinetic energy is proportional to x^2 or square of the displacement

Answer 2

Explanation:

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