Physics-(general)
NEET Physics (general) MCQ Question
Type: MCQ-numerical-Medium-Class 11
A block of mass m is attached to a spring with a spring constant k. The block is displaced by a distance x from its equilibrium position, resulting in a potential energy of 1/2 k x^2. If the block is released, what will be its speed when it passes through the equilibrium position?
A
sqrt((k/m) * x^2)
B
sqrt((2k/m) * x^2)
C
x * sqrt(k/m)
D
2x * sqrt(k/m)
Correct Answer
Option C
Detailed Explanation
At equilibrium, all potential energy is converted to kinetic energy: (1/2)kx^2 = (1/2)mv^2. Thus, v = x * sqrt(k/m).
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