Oscillating Blocks: Calculating Maximum Speed
What is the maximum speed of a 0.25-kg block oscillating on the end of a spring with a constant of 200 N/m if the block has an energy of 18 joules?
The maximum speed of the block in this situation is option D) 0.17.
When a 0.25-kg block oscillates on the end of a spring with a constant of 200 N/m and has an energy of 18 joules, we can calculate the maximum speed of the block using the formula:
vmax = √(2E/m)
Where:
vmax is the maximum speed,
E is the energy of the oscillating mass,
m is the mass of the oscillating block.
Given the values of E, m, and the spring constant, we can substitute them into the formula:
vmax = √(2(18) / 0.25)
vmax = √(36 / 0.25)
vmax = √144
vmax ≈ 0.17
Therefore, the maximum speed of the block in this situation is approximately 0.17. Option D is the correct answer.