Solution:
Initial kinetic energy is $K_i = \frac{1}{2}mv^2 = 500\text{ J}$. Work done by the retarding force is $W = \int_{20}^{30} (-0.1x)dx = -25\text{ J}$. Using work-energy theorem, final K.E. $K_f = K_i + W = 500 - 25 = 475\text{ J}$.
A block of mass $10\text{ kg}$ moving in $x$ direction with a constant speed of $10\text{ m s}^{-1}$, is subjected to a retarding force $F = -0.1x\text{ J/m}$ during its travel from $x = 20\text{ m}$ to $30\text{ m}$. Its final K.E. will be: (2015)
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