First slide

Work Energy Theorem

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

Moderate

A

x²
B

e^x
C

x
D

log_e x

Solution

$Here, a = - kx, where a is acceleration, x is displacement and k is proportionality constant. ∴ v \frac{dv}{dx} = - kx Vdv = - kxdx Integrating within proper limits, we get \int_{v_{0}}^{v} vdv = - k \int_{0}^{x} xdx {[\frac{v^{2}}{2}]}_{v_{0}}^{v} = - k {[\frac{x^{2}}{2}]}_{0}^{x} (\frac{v^{2} - {v_{0}}^{2}}{2}) = - \frac{{kx}^{2}}{2} multiplying m on both sides, we get m (\frac{v^{2} - {v_{0}}^{2}}{2}) = - \frac{{mkx}^{2}}{2} \infty x^{2} Δk = - \frac{{mkx}^{2}}{2} \propto x^{2}$

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