Comprehension: A block weighing 10 N is attached to the lower end of a vertical spring ( k = 200 N/m), the other end attached to a ceiling. The block oscillates vertically and has kinetic energy 2.0 J

Question

Comprehension: A block weighing 10 N is attached to the lower end of a vertical spring ( $k = 200$ N/m), the other end attached to a ceiling. The block oscillates vertically and has kinetic energy 2.0 J as it passes through the point where the spring is unstretched. The maximum kinetic energy of the block as it oscillates ( $g = 10$ m/s²) is:

Options:

Answer:

B

Solution:

Equilibrium extension = $10/200 = 0.05$ m. From the natural length to equilibrium: $K E_{ma x} = 2 + m g (0.05) - \frac{1}{2} k 0.05^{2} = 2 + 0.5 - 0.25 = 2.25$ J.

TestHub · Accepted Answer

Equilibrium extension = 10/200 = 0.05 m. From the natural length to equilibrium: K E ma x ​ = 2 + m g ( 0.05 ) − 2 1 ​ k 0.05 2 = 2 + 0.5 − 0.25 = 2.25 J.

Physics - SHM Question with Solution | TestHub

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