A spring of negligible mass and spring constant k is fastened on one end to a wall and attached at the other end to a block of mass m. the block is pulled back a distance d from the equilibrium position and released. At what distance from the equilibrium position is the kinetic energy of the block equal to the elastic potential energy?
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The maximum potential energy = 0.5 kd²
Kinetic energy at y + P.E at y = the total energy or the maximum P.E = 0.5 kd²
When K.E equal p.E
2.(P.E at y) = 0.5 kd²
2.* 0.5 k y² = 0.5 kd²
2. y² = d²
y = d / √2 = d / 1.414
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Kinetic energy at y + P.E at y = the total energy or the maximum P.E = 0.5 kd²
When K.E equal p.E
2.(P.E at y) = 0.5 kd²
2.* 0.5 k y² = 0.5 kd²
2. y² = d²
y = d / √2 = d / 1.414
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