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------------------------------------------------ A very large mass-spring system has a stiffness of 193 N/m and a mass of 0.44 kg. It is oriented horizontally atop a level table. The system is released with an initial compression of 9 cm and an initial speed of 4 m/s.

Part (a): What is the maximum stretch of the spring during the motion (in meters)?

To find the maximum stretch of the spring, we can use the conservation of mechanical energy and the equation for potential energy of a spring. The maximum stretch of the spring is approximately 0.41 m.

Explanation:

To find the maximum stretch of the spring, we can use the conservation of mechanical energy. The initial energy of the system is the sum of the potential energy stored in the compressed spring and the kinetic energy of the mass.

Using the equation ΔE = ΔPE + ΔKE, we can equate the change in energy to zero. Rearranging the equation gives us ΔPE = -ΔKE. We can calculate the change in kinetic energy using the equation ΔKE = (1/2)mv^2, where m is the mass of the system and v is the initial speed. Plugging in the given values, we find that the change in potential energy is -0.88 J.

To find the maximum stretch, we can use the formula for potential energy of a spring, PE = (1/2)kx^2, where k is the stiffness of the spring and x is the maximum stretch. Plugging in the known values, we can solve for x. Using the equation 0.88 J = (1/2)(193 N/m)x^2, we find that the maximum stretch of the spring is approximately 0.41 m.

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