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GP A \(60.0-\mathrm{kg}\) athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0\). What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

Step by step solution

01

Identify Interacting Objects and Their Interactions

The interacting objects here are athlete and the Earth. They interact through gravitational force.

02

Determine the Kinetic and Potential Energy at Initial Position

Given the starting speed is \(9.0 \, \mathrm{m/s}\), we can calculate the initial kinetic energy using the formula \(KE = 0.5mv^2\), where \(m = 60.0 \, \mathrm{kg}\) and \(v = 9.0 \, \mathrm{m/s}\). The gravitational potential energy at \(y = 0\) is \(0\).

03

Determine the Kinetic and Potential Energy at Maximum Height

At maximum height, the kinetic energy is zero since the athlete is momentarily stationary. The potential energy can be calculated through the formula \(PE= mgh\). Since \(m = 60 \, \mathrm{kg}\) and \(g = 9.8 \, \mathrm{m/s^2}\) we need to find \(h\).

04

Apply Conservation of Energy and Solve for Maximum Height

In terms of energy, conservation law says initial kinetic energy would be equal to final potential energy (since at maximum height kinetic energy is zero). Set them equal and solve for \(h\): \(0.5mv^2 = mgh\)

05

Find Velocity at Half the Maximum Height

At half the maximum height, the athlete has both kinetic and potential energy. If we set \(h\) as half the maximum height, we can use the conservation of energy to solve for velocity: \(0.5mv^2 = 0.5mgh + 0.5mv'^2\) where \(v'\) is the speed at half the maximum height.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy

Kinetic energy is the energy of motion. Any moving object has kinetic energy. It's calculated using the formula:\[KE = \frac{1}{2}mv^2\]where \(m\) is the mass of the object and \(v\) is its velocity. In the exercise, the athlete has a mass of 60 kg and an initial speed of 9 m/s when she starts her leap.
This means her kinetic energy at the starting point is calculated as:

• Substitute into the formula: \(KE = \frac{1}{2} \times 60 \times (9)^2 = 2430 \text{ Joules}\).

When she reaches maximum height, her speed (and thus kinetic energy) becomes zero, as she momentarily stops before falling back down. Knowing this is key to solving for her maximum height using energy principles.

Potential Energy

Potential energy is stored energy. For objects in a gravitational field, like Earth's, it's often called gravitational potential energy. It's calculated with:\[PE = mgh\]where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and \(h\) is the height above the reference point.
In our exercise, at the initial jump point, the reference height \(y = 0\), so potential energy is zero there. As the athlete rises to the maximum height, kinetic energy is converted to potential energy.

• At maximum height, all kinetic energy converts into potential energy, allowing us to use energy conservation: \(mgh = 2430 \text{ Joules} \).

This relationship lets us solve for the maximum height that the athlete achieves.

Gravitational Force

Gravitational force is the attractive force that the Earth exerts on objects, pulling them towards its center. It governs how objects interact when they are dropped or thrown - keeping them grounded or pulling them back when they try to rise into the air. The strength of this force helps define potential energy, which an object has due to its position in a gravitational field.
Gravity's pull is what decelerates the athlete as she rises, converting her kinetic energy into potential energy until she reaches her zenith and stops momentarily. At this point:

• The gravitational potential energy is at its maximum, equivalent to the initial kinetic energy.
• The formula \(mgh\) incorporates gravitational force into potential energy considerations.

Understanding gravitational force is essential to grasping how energy transitions occur during motion, such as our athlete's leap.