A 0.25 kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the amplitude of the oscillation is
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Step 1
In this case, the block is initially displaced by 0.15 m and given an initial speed of 3.0 m/s. The total energy of the system is the sum of the kinetic and potential energy. The kinetic energy is given by (1/2)mv^2 and the potential energy by (1/2)kx^2, where m Show more…
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