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A 0.40-kg mass is attached to a spring with a force constant of k = 367 N/m, and the mass–spring system is set into oscillation with an amplitude of A = 2.4 cm. Determine the following. (a) Mechanical energy of the system J Options: A. 0.552 J B. 0.275 J C. 0.110 J D. 0.689 J

User Areg
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Final answer:

The mechanical energy of the mass-spring system set into oscillation is the potential energy of the spring when it is stretched to its maximum amplitude, which is 0.552 J.The correct option is C. 0.110 J.

Step-by-step explanation:

To determine the mechanical energy of the mass-spring system, we use the formula for the energy stored in a spring undergoing simple harmonic motion (SHM), which is the sum of its potential energy at maximum compression or extension (due to the spring) and its kinetic energy at the equilibrium position (maximum speed).

Since there is no mention of friction or air resistance, we can assume the energy is conserved, and thus the mechanical energy of the system will be purely the potential energy of the spring at its maximum amplitude. The potential energy of the spring (U) can be calculated by the equation:

U = ½ kA²

where k is the spring constant, and A is the amplitude. For a spring with k = 367 N/m and A = 2.4 cm (converted to meters as 0.024 m, since we need SI units), the mechanical energy of the system is:

U = 0.5 × 367 N/m × (0.024 m)² = 0.5 × 367 × 0.024² = 0.5 × 367 × 0.000576 JAnswer Option A: 0.552 J.The correct option is C. 0.110 J.

User Fersca
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