Question

Calculate the kinetic energy and the velocity required for a 70kg pole vaulter to pass over a 5.0m high bar. Assume the vaulter’s centre of mass is initially 0.90m off the ground and reaches its maximum height at the level of the bar itself.

Answer 1

The pole vaulter needs to achieve a kinetic energy of 2774.6 J and a minimum velocity of 7.92 m/s to clear a 5.0m high bar, using the conservation of energy principle.

Step-by-step explanation:

To calculate the kinetic energy and the velocity required for a 70kg pole vaulter to pass over a 5.0m high bar, we will use the conservation of energy principle. Assuming that the initial potential energy is converted into kinetic energy at the top of the jump (maximum height), we can set up the conservation of energy equation as follows:

1. Calculate the difference in height: Δh = final height - initial height = 5.0m - 0.90m = 4.1m.
2. Calculate the potential energy at the initial position using PE = mgh, where g = 9.8 m/s² (acceleration due to gravity), m is the mass (70 kg), and h is the initial height (4.1m).
3. Set the initial potential energy equal to the final kinetic energy. KE = PE = mgh = 70kg * 9.8m/s² * 4.1m.
4. Find the kinetic energy. KE = 2774.6 J (joules).
5. Using the kinetic energy, calculate the velocity (v) using the formula KE = 1/2 mv². Solve for v to get v = √(2KE/m).
6. Calculate the velocity: v = √(2 * 2774.6J / 70kg) which results in v ≈ 7.92 m/s.

The pole vaulter will need a kinetic energy of 2774.6 J and a minimum velocity of 7.92 m/s to clear the 5.0m high bar.

Answer 2

Below

Step-by-step explanation:

The vaulter's entire KE will be converted to PE to clear the bar (5-.9) = 4.1 m above the COM.

PE required = mgh = 70 ( 9.81)(4.1) = 2815.5 J

KE required will then be 2815.5 J

(As an aside: = 1/2 mv^2 = 2815.5 = 1/2 ( 70)(v^2) shows v = 9 m/s)