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Question

A spring has natural length of 50 cm and a force constant of 2.0×103 Nm1 . A body of mass 10 kg is suspended from it in equilibrium position. If the body is pulled down, such that length of spring becomes 58 cm and released. If it executes simple harmonic motion, what is the net force on the body, when it is at it's lowermost position of oscillation ?
(Take g=10 ms2).

A
20 N
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B
40 N
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C
60 N
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D
80 N
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Solution

The correct option is C 60 N
Given that,
Natural length of the spring ln=50 cm
Force constant of the spring k=2.0×103 Nm1
Extension in the length of the spring when 10 kg mass is suspended.
kx=mg
x=10 kg×10 ms22.0×103 Nm1
x=0.05 m=5 cm
Equilibrium position of the mass is
x1=50cm+5cm
x1=55 cm
Now spring is further stretched to 58 cm
So, displacement from mean position is
x=58 cm55 cm=3 cm

Net force at lowermost position is,
Fnet=kx=2×103×3×102=60 N
Hence, c is the correct option.

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