A 0.63 kg mass rests on top of a vertical spring with a spring constant of 65 N/m. Determine the distance that the spring is compressed from its equilibrium position when the mass is at rest.

The mass will sag lower and lower and compress the spring, until the upward force of the spring is equal to the weight of the mass. When that happens, the two forces on the mass (gravity and spring) will be balanced, and the mass will stop sinking.
-- The weight of the mass is (mass) x (gravity) = (0.63) x (9.8) Newtons.
-- The distance that the spring is compressed is 1 meter / 65 newtons.
(1 meter/65 newtons) x (0.63 x 9.8) newtons = 0.095 meter = 9.5 centimeters

Answered by AL2006 | 2024-06-10

The spring compresses by approximately 9.5 centimeters when a 0.63 kg mass is placed on it. This is calculated by balancing the gravitational force acting on the mass with the spring force using Hooke's Law. The weight of the mass causes the spring to compress until equilibrium is reached.
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Answered by AL2006 | 2024-11-01