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A spring has a natural length of 50 cm and a force constant of $2 \times 10^{3} {Nm}^{- 1}$ . A body of mass 10 kg is suspended from it and the spring is stretched. If the body is pulled down to a length of 58 cm and released, it executes simple harmonic motion. What is the net force on the body when it is at its lowermost position of its oscillation?
(Take $g = 10 {ms}^{- 2}$ )

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20 N

60 N

80 N

40 N

answer is C.

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Detailed Solution

At its lowermost point, spring is stretched by 8 cm or $8 \times 10^{- 2} m .$

$∴ F_{net} = k x - m g$

$= (2 \times 10^{3} \times 8 \times 10^{- 2}) - (10 \times 10) = 60 N$

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