An adventurous 9 kg baby climbs over of its crib wall and falls towards the floor, falling a total distance of 1.5m before colliding with the floor. calculate the average force of the floor on the above baby if the baby were to fall onto a hardwood floor which only compresses about 0.001 m.

The initial potential energy of the baby at the top of the fall is given by mgh, where m is the mass (9 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (1.5 m). This potential energy is converted into kinetic energy at the bottom of the fall.

The change in kinetic energy is equal to the work done by the floor, which can be calculated as the average force multiplied by the distance over which it acts. In this case, the distance is the compression of the hardwood floor, which is 0.001 m.

Setting the initial potential energy equal to the change in kinetic energy, we can solve for the average force:

mgh = (1/2)mv^2

Average force * distance = (1/2)mv^2

Average force = (1/2)mv^2 / distance

Plugging in the values:

Average force = (1/2)(9 kg)(9.8 m/s^2)(1.5 m) / 0.001 m

Average force ≈ 6612 N

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