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(A) The retardation $a$ is proportional to the displacement $x$,so $a = -kx$,where $k$ is a positive constant.According to Newton's second law,the for

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$x^2$

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(A) The retardation $a$ is proportional to the displacement $x$,so $a = -kx$,where $k$ is a positive constant.According to Newton's second law,the force $F = ma = -mkx$.The work done by this force for a displacement $x$ is given by $W = \int_{0}^{x} F \cdot dx$.$W = \int_{0}^{x} (-mkx) \cdot dx = -mk \int_{0}^{x} x \cdot dx = -mk \left[ \frac{x^2}{2} \right]_{0}^{x} = -\frac{1}{2} mkx^2$.According to the work-energy theorem,the work done is equal to the change in kinetic energy,$W = \Delta KE = KE_f - KE_i$.Since the force is a retardation,the kinetic energy decreases,so the loss of kinetic energy is $|W| = \frac{1}{2} mkx^2$.Therefore,the loss of kinetic energy is proportional to $x^2$.

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

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