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A 50 kg Bungee jumper drops from a bridge with an initial speed of 0 m/s. The height from the bridge to the ground is 75 m. If the natural length of the bungee (i.e. spring) attached to the jumper is 1/3 of the height of the bridge, what is the minimum spring constant k that will prevent this jumper from touching the ground after jumping? (g=9.8 m/s²)

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Sure, let's calculate the minimum spring constant that will prevent the jumper from touching the ground.

Following the law of conservation of energy, the total energy E of the jumper is equal to his potential energy plus his kinetic energy.

Since the bungee jumper is not moving initially, his kinetic energy (which is calculated by 1/2 * mass * (velocity)^2) is zero. The potential energy due to gravity can be calculated by mass * gravity * height. So in this case, the potential energy is equal to the total energy, resulting in potential energy being equal to 50 kg * 9.8 m/s² * 75 m.

The total energy of the jumper then gets converted entirely into the potential energy of the spring when he is at the lowest position, just before touching the ground. We can represent this as 1/2 * k * (distance stretched by the spring)^2, where k is the spring constant we're trying to find. Setting this equal to the potential energy allows us to solve for k.

However, with bungee jumping, the spring is not getting stretched from its natural length to zero. It's getting stretched from its natural length to the total height from the bridge to the ground. Therefore, we consider the distance over which the spring is stretching to be the height minus the natural length of the spring.

The natural length of the spring is given as a third of the height, so we subtract this from the total height to get the stretched length of the spring.

Plugging this value into our formula and solving for k, we find that the minimum spring constant that will prevent the jumper from touching the ground is approximately 29.4 N/m.

User Marat Batalandabad
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