A block of mass 10kg is moving in x direction with a constant speed of 10 m/s. It is subjected to a retarding force F = -0.1 x J/m during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be: (a) 475 J (b) 450 J (c) 275 J (d) 250 J

Applying work energy theorem

W = (KE)f – (KE)i

\((KE)_{i}=\frac{1}{2}mv^{2} = \frac{1}{2}\times 10\times 10^{2}\)

On calculation, we get,

(KE)i = 500 J

\(W = \int_{20}^{30}F.dx = \int_{20}^{30}-0.1 x dx\\= -0.1\left [ \frac{x^{2}}{2} \right ]_{20}^{30}\\=\frac{-0.1}{2}\left ( 900 – 400 \right )\\-0.1\times \frac{500}{2}\)

We get,

W = -25 J

W = (KE)f – (KE)i

Substituting the values

-25 J = (KE)f – 500 J

(KE)f = 500 J – 25 J

We get,

(KE)f = 475 J

Hence, the correct option is (a)

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