Chapter 8: Problem 13
(II) In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass \(2.10 \mathrm{~m}\) and cross the bar with a speed of \(0.70 \mathrm{~m} / \mathrm{s} ?\)
Short Answer
Step by step solution
Understanding the Energy Conversion
Writing the Equations for Energy
Applying the Conservation of Energy
Simplifying and Solving for Initial Speed
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy
- The more speed an athlete has, the more kinetic energy is generated.
- A higher kinetic energy means the athlete can reach greater heights.
Potential Energy
The potential energy is especially relevant when discussing objects in a gravitational field, such as when a high jumper reaches their highest point.
- Potential energy depends on the height of the object. The higher an object is, the more potential energy it has.
- In physics, potential energy is often discussed alongside kinetic energy in energy conservation problems.
Gravitational Potential Energy
It is calculated using the formula \[ PE = mgh \], where:
- \( m \) is the mass of the object.
- \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
- \( h \) is the height above the ground.
- The higher the athlete jumps, the greater the gravitational potential energy they accumulate.
- This energy is vital for overcoming the gravitational pull that wants to bring the jumper back to the ground.
High Jump
- Starting with a run allows athletes to gather significant kinetic energy.
- At takeoff, they convert this kinetic energy into potential energy, helping them rise.
- Once they clear the bar, some kinetic energy remains, allowing the athlete to continue their motion and safely land.