Force constant of a weightless spring is 16N/m. A body of mass 1.0kg suspended from it is pulled down through 5cm from its mean position and then released. The maximum kinetic energy of the body will be
A
2×10−2J
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B
4×10−2J
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C
8×10−2J
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D
16×10−2J
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Solution
The correct option is A2×10−2J Let us assume , spring- mass system is initially in equilibrium at x=0. Let this point be the reference point about which the body executes SHM.
Given,
Mass of the body m=1.0kg
Elongation of spring/Amplitude of SHM A=5cm
Maximum kinetic energy of the body attached to the spring is given by Kmax=12kA2 =12×16×(5×10−2)2=2×10−2J
Thus, option (a) is the correct answer.