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The length of a smooth inclined plane of inclination $30^{0}$ is $10 m$ . The work done in moving a $20 k g$ mass from the bottom of the inclined plane to top is

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245 J

490J

980J

500J

answer is C.

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Detailed Solution

$w = (m g \sin θ) s = 20 \times 9.8 \times \sin 30^{0} \times 10$

$= 100 \times 9.8 = 980 J$

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