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Chapter 13: Problem 10

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?

Short Answer

Expert verified
The rock loses \(4.7 * 10^6 J\) of mechanical energy due to friction.

Step by step solution

01

Calculate initial potential energy

Calculate the initial potential energy that the rock has at the top of the hill using the equation \(PE_i = mgh\), where \(m\) is the mass of the rock, \(g\) is the acceleration due to gravity (approximated as \(9.8 m/s^2\)), and \(h\) is the height of the hill. Plugging the given values this results in \(PE_i = 524 kg * 9.8 m/s^2 * 292 m = 1.5 * 10^7 J\).
02

Calculate final kinetic energy

Calculate the final kinetic energy that the rock has at the bottom of the hill using the equation \(KE_f = \frac{1}{2}mv^2\), where \(m\) is the mass of the rock and \(v\) is the speed. Plugging the given values this results in \(KE_f = \frac{1}{2} * 524 kg * (62.6 m/s)^2 = 1.03 * 10^7 J\).
03

Calculate energy loss due to friction

The mechanical energy lost due to friction is the difference between the initial potential energy and the final kinetic energy. Therefore, the energy lost due to friction is calculated as \(E_f = PE_i - KE_f = 1.5 * 10^7 J - 1.03 * 10^7 J = 4.7 * 10^6 J\).

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Most popular questions from this chapter

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.

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 \(3 \mathrm{ft}\) ?

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.

The National Transportation Safety Board is testing the crashworthiness of a new car. The \(2340-\mathrm{kg}\) vehicle is driven at \(12.6 \mathrm{~km} / \mathrm{h}\) into an abutment. During impact, the center of mass of the car moves forward \(64.0 \mathrm{~cm} ;\) the abutment is compressed by \(8.30 \mathrm{~cm}\). Ignore friction between the car and the road. (a) Find the force, assumed constant, exerted by the abutment on the car. (b) By how much does the internal energy of the car increase?

A ball of mass \(12.2 \mathrm{~g}\) is dropped from rest at a height of \(76 \mathrm{~cm}\) above the surface of oil that fills a barrel to a depth of \(55 \mathrm{~cm}\). The ball reaches the bottom of the barrel with a speed of \(1.48 \mathrm{~m} / \mathrm{s}\). (a) Neglecting air resistance, find the speed of the ball when it enters the oil. \((b)\) What is the change in the internal energy of the system of ball + oil?
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