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A force of 400 Newtons stretches a spring 2 meters. A mass of 50 kg is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10m/s. Find the equation of motion.

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Final answer:

The equation of motion for the mass attached to the spring, using SHM with the calculated spring constant and the given initial conditions, is x(t) = -5 cos(2t).

Step-by-step explanation:

To find the equation of motion for the given system, we must first calculate the spring constant k. The force required to stretch the spring 2 meters is 400 Newtons, therefore k = F/x = 400 N/2 m = 200 N/m. With this spring constant and a mass of 50 kg, we are dealing with a situation of simple harmonic motion (SHM).

The general equation for SHM is x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Angular frequency ω = sqrt(k/m), so we have ω = sqrt(200/50) = 2 rad/s. Since the object starts from the equilibrium position but with an upward velocity, we can deduce that the phase constant φ is zero.

Knowing the initial velocity is 10 m/s, we differentiate the SHM equation to find velocity: v(t) = -ωA sin(ωt + φ). At t=0, v(0) = 10 m/s = -ωA sin(φ), which simplifies to 10 m/s = -2A since φ = 0. Thus, A = -5 m. Finally, we have the equation of motion: x(t) = -5 cos(2t).

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