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.

Question

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.

Answer 1

To determine how far the spring would stretch downward once the baby is placed in the seat, we need to equate the gravitational force on the baby (Fg) to the elastic force exerted by the spring on the baby (Fe).

The gravitational force on the baby (Fg) is given by the formula:

Fg = m * g

where m is the mass of the baby and g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, the mass of the baby (m) is given as 11 kg. Therefore, the gravitational force on the baby is:

Fg = 11 kg * 9.8 m/s^2
≈ 107.8 N

According to Hooke's law, the elastic force exerted by the spring (Fe) is given by:

Fe = -k * x

where k is the spring constant and x is the displacement from equilibrium.

In this case, the spring constant (k) is given as 500 N/m. We want to find the displacement (x).

Now we can equate the gravitational force (Fg) to the elastic force (Fe):

Fg = Fe

Substituting the given values:

107.8 N = -500 N/m * x

To solve for x, we divide both sides of the equation by -500 N/m:

x = 107.8 N / -500 N/m
≈ -0.2164 m

The negative sign indicates that the displacement is downward (opposite to the upward direction).

Therefore, the spring would stretch downward by approximately 0.2164 meters once the baby is placed in the seat.