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Q.

A block of mass 10 kg, moving in x direction with a constant speed of 10 m/s, is subject to a retarding force F = (0.1)x N during its travel from x = 20 m to 30 m. Its final KE will be

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a

475 J

b

275 J

c

250 J

d

450 J

answer is D.

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Detailed Solution

From, F = ma
$a = \frac{F}{m} = \frac{0.1 x}{10} = 0.01 x = V \frac{d V}{d x}$
So, $\int_{v_{1}}^{v_{2}} v d V = \int_{20}^{30} \frac{x}{100} d x$
${- \frac{V^{2}}{2}|}_{V_{1}}^{V_{2}} = {\frac{x^{2}}{200}|}_{20}^{30} = \frac{30 \times 30}{200} - \frac{20 \times 20}{200}$
$= 4.5 - 2 = 2.5$
$\frac{1}{2} m (V_{2}^{2} - V_{1}^{2}) = 10 \times 2.5 J = - 25 J$
Final K.E.
$= \frac{1}{2} m v_{2}^{2} = \frac{1}{2} m v_{1}^{2} - 25 = \frac{1}{2} \times 10 \times 10 \times 10 - 25$
$= 500 - 25 = 475 J$

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