Problem 13 - Physics for Scientists and Engineers with Modern Physics

Question

(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 \(0.70 \mathrm{~m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

The minimum speed required is approximately 6.46 m/s.

Step 1: Understanding the Energy Conversion

The high jumper's kinetic energy (KE) at takeoff is partially converted into gravitational potential energy (PE) at the highest point. At the bar, the jumper still has some kinetic energy due to their speed. We start with the energy conservation equation: \[\text{Initial KE} = \text{Final KE} + \text{PE}\]

Step 2: Writing the Equations for Energy

Initially, the kinetic energy is given by:\[ KE_i = \frac{1}{2}m v_i^2 \]At the bar, the athlete has both gravitational potential energy and kinetic energy. The potential energy is:\[ PE_f = mgh \]The final kinetic energy at the bar is:\[ KE_f = \frac{1}{2}m v_f^2 \]Where:- \(m\) is the mass of the athlete (which will cancel out)- \(v_i\) is the initial speed- \(v_f = 0.70 \text{ m/s}\), final speed at the bar- \(h = 2.10 \text{ m}\), height of the bar- \(g = 9.81 \text{ m/s}^2\), acceleration due to gravity

Step 3: Applying the Conservation of Energy

According to the conservation of energy, the initial kinetic energy equals the sum of final kinetic energy and gravitational potential energy:\[\frac{1}{2}m v_i^2 = \frac{1}{2}m v_f^2 + mgh\]

Step 4: Simplifying and Solving for Initial Speed

Since the mass \(m\) appears in all terms, it cancels out:\[\frac{1}{2}v_i^2 = \frac{1}{2}v_f^2 + gh\]Plugging in the values for \(v_f\), \(g\), and \(h\):\[\frac{1}{2}v_i^2 = \frac{1}{2}(0.70)^2 + (9.81)(2.10)\]Calculate the right side:\[\frac{1}{2}(0.49) + 20.601 = 0.245 + 20.601 = 20.846\]Thus,\[\frac{1}{2}v_i^2 = 20.846\]Multiply by 2:\[v_i^2 = 41.692\]Taking the square root:\[v_i = \sqrt{41.692} \approx 6.46 \text{ m/s}\]

Step 5: Conclusion

The minimum speed the athlete must leave the ground to reach the necessary height and cross the bar is approximately \(6.46 \text{ m/s}\).

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It's a fundamental concept in physics that applies to moving objects. The kinetic energy (\( KE \)) of an object with mass (\( m \)) and velocity (\( v \)) is calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]This energy type is crucial in many physical phenomena, including sports like the high jump. When a high jumper takes off, they use their speed to gain kinetic energy, which helps them overcome gravity and propel their body upward.

• The more speed an athlete has, the more kinetic energy is generated.
• A higher kinetic energy means the athlete can reach greater heights.

The relationship between speed and kinetic energy is quadratic, meaning that doubling the speed results in four times the kinetic energy. Understanding this energy helps explain how athletes are able to jump high bars.

Potential Energy

Potential energy is stored energy in an object due to its position or state. It has the potential to do work when released. In the case of a high jumper, the potential energy comes into play as the athlete rises. This energy type contrasts with kinetic energy, which involves motion.
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.

Understanding potential energy is key to grasping how high jumpers convert their kinetic energy (from running) into the energy necessary to lift their bodies over the bar.

Gravitational Potential Energy

Gravitational potential energy is a type of potential energy related to the height of an object in a gravitational field. When athletes perform a high jump, they aim to increase this form of energy while defying gravity.
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.

In the context of a high jump, this means:

• 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.

Understanding gravitational potential energy is vital for explaining how energy conversions take place at the peak of a jump.

High Jump

The high jump is a classic track and field event where athletes use a running start to propel themselves over a horizontal bar. It's a great real-world example of energy conservation, where kinetic energy from running is converted into gravitational potential energy to achieve height. The event demonstrates numerous physics concepts in action:

• 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.

The high jump elegantly illustrates the principles of energy conversion and conservation, highlighting how athletes harness and optimize their body's energy to achieve greater heights.