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A 6kg block starts from rest against a spring with a spring constant of 500N/m that is compressed by a distance of 2m. The ground is frictionless. What will be the velocity of the block leaving the spring?

Sagot :

The velocity of the block leaving the spring is determined as 18.26 m/s.

Velocity of the block

The velocity of the block leaving the spring is calculated from the principle of conservation of energy.

K.E = Ux

¹/₂mv² = ¹/₂kx²

v² = kx²/m

v² = (500 x 2²)/6

v² = 333.33

v = √333.33

v = 18.26 m/s

Thus, the velocity of the block leaving the spring is determined as 18.26 m/s.

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