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Chapter 13: Problem 5

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?

Short Answer

Expert verified
The speed of the ball when it enters the oil is determined in Step 1. The change in the internal energy of the system of ball+oil is then calculated in Step 3 using the kinetic energies at the starting and end points.

Step by step solution

01

Calculate the initial speed

Use the equation of motion \(v^2 = u^2 + 2gs\) to find the speed 'v' of the ball when it enters the oil. Here, 'u' is the initial speed which is 0 (since the ball is dropped), 'g' is the acceleration due to gravity (9.81 m/s²), and 's' is the distance the ball travels before hitting the oil (in meters). It's important to change the distance from cm to m by dividing by 100, so \(s = 76/100 = 0.76m\). Now, let's substitute these values into the equation to find 'v'.
02

Calculate the change in internal energy

To find the change in internal energy of the system of ball + oil, first we will need to calculate the initial and final kinetic energy. The initial kinetic energy of the system, considering the ball about to enter the oil, is given by \(KE_i = 0.5 * m * v^2\), where m is the mass of the ball and v is the speed calculated in the previous step. The final kinetic energy of the system when the ball hits the bottom of the barrel with a speed of 1.48 m/s is given by \(KE_f = 0.5 * m * (1.48)^2\). The change in internal energy is then given by \(\Delta E = KE_f - KE_i\).
03

Calculate the final values

Use the values of mass \(m = 12.2 / 1000 = 0.0122 kg\) (converted from g to kg), and the speeds we calculated for 'v' in step one and from the problem statement to find the values of KE_i and KE_f. Subtract KE_i from KE_f to get the change in internal energy.

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Most popular questions from this chapter

An electron, mass \(m\), collides head-on with an atom, mass \(M\), initially at rest. As a result of the collision, a characteristic amount of energy \(E\) is stored internally in the atom. What is the minimum initial speed \(v_{0}\) that the electron must have? (Hint: Conservation principles lead to a quadratic equation for the final electron speed \(v\) and a quadratic equation for the final atom speed \(V\). The minimum value, \(v_{0}\), follows from the requirement that the radical in the solutions for \(v\) and \(V\) be real.)

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 steel ball of mass \(0.514 \mathrm{~kg}\) is fastened to a cord \(68.7 \mathrm{~cm}\) long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a \(2.63-\mathrm{kg}\) steel block initially at rest on a frictionless surface (Fig. \(13-18\) ). On collision, one-half the mechanical kinetic energy is converted to internal energy and sound energy. Find the final speeds.

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.

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.
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