A girl with a mass of 83 kg is climbing in a tree 20 m above the ground. If she lands on a trampoline with a constant of 312 N/m, how far will the trampoline stretch and how fast will she be moving just prior to landing on it?
V^2 = Vo^2 + 2g*d. Solve for V.
To determine how far the trampoline will stretch and how fast the girl will be moving prior to landing, we can use the principles of conservation of energy and Hooke's Law.
1. First, let's calculate the potential energy the girl has when she's 20 meters above the ground. The potential energy is given by the formula: potential energy = mass * acceleration due to gravity * height.
Potential energy = 83 kg * 9.8 m/s^2 * 20 m = 16,264 J
2. Next, we know that potential energy is converted into kinetic energy as the girl falls. The kinetic energy is given by the formula: kinetic energy = (1/2) * mass * velocity^2. Since the girl starts from rest, her initial kinetic energy is zero.
16,264 J = (1/2) * 83 kg * velocity^2
3. We can rearrange the equation to solve for the velocity: velocity = sqrt((2 * potential energy) / mass).
velocity = sqrt((2 * 16,264 J) / 83 kg) = 14.69 m/s (rounded to two decimal places)
4. Now, using Hooke's Law, we can calculate how far the trampoline will stretch. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
The force exerted by the trampoline is given by the formula: force = spring constant * displacement.
By rearranging the equation, we get: displacement = force / spring constant.
force = mass * acceleration = 83 kg * 9.8 m/s^2 = 813.4 N
displacement = 813.4 N / 312 N/m = 2.61 m (rounded to two decimal places)
Therefore, the trampoline will stretch by approximately 2.61 meters, and the girl will be moving at a velocity of 14.69 m/s just prior to landing on it.