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A block of mass 10 kg, moving in x-direction with a speed of 10 ms^–1, is subject to a retarding force F = 2.5 N during its travel from x = 20 m to 30 m. Its final Kinetic Energy will be

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450 J

250 J

475 J

275 J

answer is D.

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Detailed Solution

Work-energy theorem,

$\frac{1}{2} m (V_{2}^{2} - V_{1}^{2}) = F \times S \Rightarrow Final K . E . = \frac{1}{2} {mv}_{2}^{2} = \frac{1}{2} {mv}_{1}^{2} - F \times S = \frac{1}{2} \times 10 \times 10 \times 10 - 25 = 500 - 25 = 475 J$

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