Posted by **Anonymous** on Sunday, February 20, 2011 at 11:44am.

Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, F, the distance from equilibrium the end of the spring is displaced, x, and a number k called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium: F=-kx. The negative sign indicates that the force that the spring exerts and its displacement have opposite directions.

So, here is the funny question for you. You create a baby bouncer using an elastic cord (consider it to be a spring). Use Hooke's law to determine how far the spring would stretch downward once the baby is placed in the seat, if the spring constant k = 500 N/m, and the baby has a mass m = 11 kg. Hint: weight of the baby (gravitational force on it) is the force exerted on the spring. In equilibrium (when baby is at rest) it's magnitude becomes equal to the elastic force exerted by the spring on baby: F_g = F_e.

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