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Chapter 13: Problem 13

A ball loses \(15.0 \%\) of its kinetic energy when it bounces back from a concrete walk. With what speed must you throw it vertically down from a height of \(12.4 \mathrm{~m}\) to have it bounce back to that same height? Ignore air resistance.

Short Answer

Expert verified
Following these steps should result in the solution to the problem. Plug the known values into the equation derived from the law of energy conversation and solve for the initial speed required for the ball to bounce back to the original height.

Step by step solution

01

Analyze the Energy Loss

The ball loses \(15.0\%\) of its kinetic energy when it bounces back. Therefore, the loss fraction will be \(15.0\% / 100 = 0.15\).
02

Apply the Law of Conservation of Energy

Let's denote the initial speed as \( v \). The potential energy \( mgh \) right before hitting the ground is converted into kinetic energy \(0.5 * m * v^2 \) when it hits the ground. After hitting, the ball retains \( (1 - 0.15) = 0.85 \) or \(85\%\) of its kinetic energy. This fraction of its energy is enough to take it back to the original height. Therefore, we can write the energy conservation equation as \( mgh = 0.85 * 0.5 * m * v^2 \). The mass \( m \) of the ball cancels out.
03

Evaluation

The height \( h \) is given as \( 12.4m \) and gravitational acceleration \( g \) as \( 9.8m/s^2 \). Substituting the known values and solving for \( v \), we get \( v = \sqrt{2gh/0.85} \).
04

Calculation

Substituting \( h = 12.4m \) and \( g = 9.8m/s^2 \) into the formula and performing the calculation, we obtain the initial speed \( v \) required for the ball to bounce back to the original height.

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Most popular questions from this chapter

A rubber ball dropped from a height of exactly \(6 \mathrm{ft}\) bounces (hits the floor) several times, losing \(10 \%\) of its kinetic energy each bounce. After how many bounces will the ball subsequently not rise above \(3 \mathrm{ft}\) ?

A \(25.3-\mathrm{kg}\) bear slides, from rest, \(12.2 \mathrm{~m}\) down a lodge- pole pine tree, moving with a speed of \(5.56 \mathrm{~m} / \mathrm{s}\) at the bottom. \((a)\) What is the initial potential energy of the bear? \((b)\) Find the kinetic energy of the bear at the bottom. (c) Assuming no other energy transfers, find the change in internal energy of the bear and tree.

Let the total energy of a system of \(N\) particles be measured in an arbitrary frame of reference, such that \(K=\Sigma \frac{1}{2} m_{n} v_{n}^{2} .\) In the center-of-mass reference frame, the velocities are \(v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}\), where \(v_{\mathrm{cm}}\) is the velocity of the center of mass relative to the original frame of reference. Keeping in mind that \(v_{n}^{2}=\overrightarrow{\mathbf{v}}_{n} \cdot \overrightarrow{\mathbf{v}}_{n}\), show that the kinetic energy can be written $$K=K_{\mathrm{int}}+K_{\mathrm{cm}}$$ where \(K_{\text {int }}=\Sigma \frac{1}{2} m_{n} v_{n}^{\prime 2}\) and \(K_{\mathrm{cm}}=\frac{1}{2} M v_{\mathrm{cm}}^{2} .\) This demonstrates that the kinetic energy of a system of particles can be divided into an internal term and a center-of-mass term. The internal kinetic energy is measured in a frame of reference in which the center of mass is at rest; for example, the random motions of the molecules of gas in a container at rest are responsible for its internal translational kinetic energy.

During a rockslide, a \(524-\mathrm{kg}\) rock slides from rest down a hill slope that is \(488 \mathrm{~m}\) long and \(292 \mathrm{~m}\) high. The speed of the rock as it reaches the bottom of the hill is \(62.6 \mathrm{~m} / \mathrm{s}\). How much mechanical energy does the rock lose in the slide due to friction?

A river descends \(15 \mathrm{~m}\) in passing through rapids. The speed of the water is \(3.2 \mathrm{~m} / \mathrm{s}\) upon entering the rapids and is \(13 \mathrm{~m} / \mathrm{s}\) as it leaves. What percentage of the potential energy lost by the water in traversing the rapids appears as kinetic energy of water downstream? What happens to the rest of the energy?
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