Physics-Simple Harmonic Motion
NEET Physics Simple Harmonic Motion MCQ Question
Type: MCQ-numerical-Easy-Class 11
A block attached to a spring has a total mechanical energy of 0.25 J. If the spring constant is 50 N/m and the block is at a displacement of 5 cm from the equilibrium position, what is the kinetic energy of the block at this position?
A
0.19 J
B
0.06 J
C
0.25 J
D
0.12 J
Correct Answer
Option A
Detailed Explanation
The total energy of the block is the sum of kinetic and potential energy. With the spring constant k = 50 N/m and displacement x = 0.05 m, potential energy is 0.0625 J. Thus, kinetic energy is 0.25 J - 0.0625 J = 0.19 J.
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