Energy in Our Daily Lives PDF

## Energy in Our Daily Lives ### What We Will Learn - Kinetic energy is the energy associated with the motion of an object. - Work is energy transferred to an object or from an object as the result of the action of an external force. - Positive work transfers energy to the object, and negative wor...

## Energy in Our Daily Lives ### What We Will Learn - Kinetic energy is the energy associated with the motion of an object. - Work is energy transferred to an object or from an object as the result of the action of an external force. - Positive work transfers energy to the object, and negative work transfers energy from the object. - Work is the scalar product of the force vector and the displacement vector. - The change in kinetic energy due to applied forces is equal to the work done by the forces. - Power is the rate at which work is done. - The power provided by a force acting on an object is the scalar product of the velocity vector of the object and the force vector. ### 5.1 Energy in Our Daily Lives No physical quantity has a greater importance in our daily lives than energy. Energy consumption, energy efficiency, and energy "production" are of the utmost economic importance and are the focus of heated discussions about national policies and international agreements. (The word production is in quotes because energy is not produced but rather is converted from a less usable form to a more usable form.) Energy also has an important role in each individual's daily routine: energy intake through food calories and energy consumption through cellular processes, activities, work, and exercise. Weight loss or weight gain is ultimately due to an imbalance between energy intake and use. Energy has many forms and requires several different approaches to cover completely. Thus, energy is a recurring theme throughout this book. We start in this chapter and the next by investigating forms of mechanical energy: kinetic energy and potential energy. But as we progress through the topics in this book you will see that other forms of energy are playing very important roles as well. Thermal energy is one of the central pillars of thermodynamics. Chemical energy is stored in chemical compounds, and chemical reactions can either consume energy from the environment (endothermic reactions) or yield usable energy to the surroundings (exothermic reactions). Our petroleum economy makes use of chemical energy and its conversion to mechanical energy and heat, which is another form of energy (or energy transfer). Later, we will see that electromagnetic radiation contains energy. This energy is the basis for one renewable form of energy - solar energy. Almost all other renewable energy sources on Earth can be traced back to solar energy. Solar energy is responsible for the wind that drives large wind turbines. The Sun's radiation is also responsible for evaporating water from the Earth's surface and moving it into the clouds, from which it falls down as rain and eventually joins rivers that can be dammed to extract energy. Biomass, another renewable energy resource, depends on the ability of plants and animals to store solar energy during their metabolic and growth processes. ### 5.2 Kinetic Energy The first kind of energy we will consider is the energy associated with the motion of a moving object: kinetic energy. Kinetic energy is defined as one half the product of a moving object's mass and the square of its speed: $K = \frac{1}{2}mv^2$ Note that the velocity is squared. Also note that kinetic energy, like work, is a scalar quantity. Because it is the product of mass and the square of a velocity, the units of kinetic energy are in $kgm^2/s^2$. Kinetic energy is an important quantity, and it has its own SI unit, the joule (J). The SI unit for force is the Newton (N) = $kg m/s^2$, and we can make a useful reminder: Energy unit: 1 J = 1 Nm = 1 $kgm^2/s^2$. Let's look at a few sample energy values to get a feeling for the size of the joule. A car of mass 1310 kg being driven at the speed of 55 mph (24.6 m/s) has a kinetic energy of: K = (1/2)(1310 kg)(24.6 m/s)$^2$ = 4.0 x 10$^5$ J The mass of the Earth is 6.0 x 10$^{24}$ kg, and it orbits the Sun with a speed of 30 x 10$^3$ m/s. The kinetic energy associated with this motion is 2.7 x 10$^{33}$ J. A person of mass 64.10 kg jogging at 3.50 m/s has a kinetic energy of 397 J, and a baseball (mass of 5.25 ounces avoirdupois = 0.142 kg) thrown at 50 mph (22.3 m/s) has a kinetic energy of 91.1 J. On the atomic scale, the average kinetic energy of an air molecule is 6 x 10$^{-21}$ J. The typical magnitudes of kinetic energies of some moving objects are presented in Figure 5.5. You can see from these examples that the range of energies involved in physical processes is tremendously large. Some other frequently used energy units are the electron volt (eV), the food calorie (Cal), and the megaton of TNT (Mt): 1 eV = 1.602 x 10$^{-19}$ J 1 Cal = 4186 J 1 Mt = 4.18 x 10$^{15}$ J On the atomic scale, 1 electron-volt (eV) is the kinetic energy that an electron gains when accelerated by an electric potential of 1 volt. The energy content of the food we eat is usually (and mistakenly) given in terms of calories but should be given in food calories. As we'll see when we study thermodynamics, 1 food calorie is equal to 1 kilocalorie. A nice round number to remember is that about 10 MJ (or 2500 food calories) of energy is stored in the food we eat each day. On a larger scale, 1 Mt is the energy released by exploding 1 million metric tons of the explosive TNT, an energy release achieved only by nuclear weapons or by catastrophic natural events such as the impact of a large asteroid. For comparison, in 2007, the total energy consumed by humans worldwide was 1.5 x 10$^{20}$ J. ### 5.3 Work In Example 5.1, the vase started at rest and had no energy. But before it was released, it had a potential energy of 306.6 J. The greater the height from which the vase is released, the greater the speed the vase will attain in falling a certain distance, and therefore the greater its kinetic energy. In other words, as we found in Example 5.1, the kinetic energy of the vase depends on the height from which it was released. The quantity that determines the vase's kinetic energy and therefore gives it its kinetic energy, we call work. We can also see from the calculation we performed in Example 5.1 that the kinetic energy also depends linearly (linearly means proportional) on the magnitude of the gravitational force. Doubling the mass of the vase would double the gravitational force acting on it, and thus double its kinetic energy. **Definition** Work is energy transferred to an object or from an object as the result of the action of a force. Work is defined as the scalar product of the force vector and the displacement vector: $W=F \cdot \Delta r$ The units of work are joules (J), the same as the units of energy. The vase gained kinetic energy, and work was done on the vase by the gravitational force. Before it was released, the vase had a potential energy. The energy associated with gravity is stored in the vase as potential energy, and the gravitational force works to convert that potential energy into kinetic energy by doing work on the vase. When the kinetic energy of a moving object is zero (i.e., the object is at rest), the force acting on it is not necessarily zero, but the work done by the force is zero. We'll discuss this more in connection with the work associated with a force that is perpendicular to the motion. ### 5.4 Work Done by a Constant Force Suppose we let the vase of Example 5.1 fall from rest, along a frictionless inclined plane that has an angle $\theta$ with respect to the horizontal plane (Fig. 5.8). We will neglect the friction force, but we will include the work done by gravity. As we discussed in Chapter 3, in the absence of friction, the acceleration along the plane is given by $a = gsin\theta$, and the force acting on the vase is $F = mg sin\theta$. From the definition given in the previous section, the work done by the constant gravitational force vector is given by $W = mg sin\theta \Delta r$, because the gravitational force vector and the displacement vector are parallel, and we have $W= F\cdot \Delta r = F\Delta r$. We can determine the kinetic energy the vase will have in this situation as a function of the displacement, $\Delta r$. Conversely, we can determine the work done by the gravitational force by using the initial and final velocities, the displacement, and the acceleration, which we obtained for one-dimensional motion in Chapter 3. To simplify the calculation by using the relationship between speed, displacement, and acceleration determined in Chapter 3, we have: $v^2 = v_0^2 + 2a\Delta r$, where the initial speed is zero, $v_0 = 0$. Then, $v^2 = 2a\Delta r$ and from the definition of kinetic energy, we have $K = 1/2mv^2 = ma\Delta r$. In other words, in this situation, the work that is done by the force and the change in kinetic energy are both equal to $ma\Delta r = mg sin\theta \Delta r$, which agrees with our results in Example 5.1. When we use $K = mv^2$ = $ma \Delta r$ to calculate the work that is done by a force that is not in the same direction as the displacement, we are only considering the component of the velocity, $ v^2$, that is in the same direction as the force. In other words, the total kinetic energy depends on both the horizontal and vertical components of the velocity. When we calculate the work done by a force, we are only considering the component of the velocity that is in the direction of the force. Let's look at a few limiting cases for the equation $W = mg sin\theta \Delta r$: - When the $ \Delta r$ is positive, the vase falls down the incline ($W= mg\sin\theta \Delta r$), increasing the kinetic energy of the vase. - For $ \theta = 90^{\circ}$ (i.e., the vase moves in the vertical direction downward), we can think of $ \theta$ as a component of the horizontal force since our coordinate system is tilted. In this case, the work done by the force also increases the vase's energy.&#x20; ### 5.5 Work Done by a Variable Force When the force acting on an object is variable, like the force resulting from a spring stretched or compressed, we can determine the work done by that force in stretching or compressing the spring. We also can determine the work done by the force of friction, which we will examine in Chapter 6. However, first we need to understand the concept of work done by a variable force.&#x20; In other words, the net work done by the net force is equal to the sum of the work done by the individual forces acting on the object. Strictly speaking, from what we'll learn in Chapter 6, this statement is only valid for conservative forces, but we will only consider conservative forces for now so we can simplify the notation. **One-Dimensional Case** - In a one-dimensional case, the equation for the work done by a variable force is $W = \int_{x_1}^{x_2} F(x) dx$.&#x20; - The $x_1$ and $x_2$ represent the initial and final positions of the object, and $F(x)$ is the force that is acting on the object.&#x20; - If the $x_1$ and $x_2$ have the same sign, the work done will be positive, and if the $x_1$ and $x_2$ have different signs, the work done will be negative. **Work-Kinetic Energy Theorem** The relationship between the kinetic energy of an object and the work done by the forces acting on it, called the **work-kinetic energy theorem**, is expressed formally as: $\Delta K = K_f - K_i = W$ Here, $K_f$ is the kinetic energy that an object has after work $W$ has been done on it, and $K_i$ is the kinetic energy before the work is done. The definitions of $W$ and $K$ are such that this equation states that the change in kinetic energy is equal to the work done on the object by the net force. We can use Equation 5.7 to solve for $K_f$ or for $K_i$: $K_f= K_i + W$ $K_i= K_f -W$ By definition, the kinetic energy cannot be less than zero. $K_i$ = 0 if all the object's kinetic energy is elastic potential energy. $K_i$ ≠ 0 if it has a constant amount of kinetic energy. The work-kinetic energy theorem implies that we have to consider all the work done by all the forces acting on the object to calculate the change in kinetic energy. We will discuss this point further in Chapter 6. **Work Done by the Gravitational Force** With the work-kinetic energy theorem as a theoretical tool, we can now take another look at the problem of an object falling under the force of gravity. This is the same situation we discussed in Example 5.1. On the way down, the work done by the gravitational force on the object is $W = \int_{y_1}^{y_2} F_g dy = mg (y_2 - y_1)$. Where $y_2$ is the final position and $y_1$ is the initial position of the object. The force of gravity, $F_g = mg$, acts in the same direction as the displacement, $y_2 - y_1$, resulting in a positive scalar product and therefore positive work. The situation is illustrated in Figure 5.10 a. Since the work is positive, the gravitational force increases the kinetic energy of the object. We can reverse the situation and toss the object vertically upward, making it a projectile and giving it an initial kinetic energy. This kinetic energy will decrease until the projectile reaches the top of its trajectory. During this time, the displacement vector $\Delta r$ points up, in the opposite direction to the force of gravity (Fig. 5.10 b). Thus, the work done by the gravitational force during the object's upward motion is: $W = \overrightarrow{F_g} \cdot \overrightarrow{\Delta r} = mg \Delta r$ Therefore, the work done by the gravitational force reduces the kinetic energy of the object during its upward motion. This conclusion is consistent with the general formula for work done by a constant force, $\overrightarrow{F} \cdot \overrightarrow{\Delta r}$, because the displacement (pointing upward) of the object and the gravitational force (pointing downward) are in opposite directions. **Work Done in Lifting and Lowering an Object** Now let's consider the situation in which a vertical external force is applied to an object, for example, by attaching the object to a rope and lifting it up or lowering it down. The work done by the external force in lifting the object a distance $\Delta r$ is $W = (F - mg)\Delta r$, where $F$ is the magnitude of the external force. This equation takes into account the fact that when $F$ > $mg$, the work done by the external force is positive, and when $F$ < $mg$, the work done by the external force is negative. This is illustrated in Figure 5.11. We can break down the work done by the external force into two parts: the work done against gravity and the work done to change the kinetic energy of the object. **Lifting with a Pulley** Let’s look at an example of a pulley. If a box is lifted a distance $\Delta h$ by pulling on a rope that is attached to a pulley, the work done by the forces is: - $W_{ext} = F_{ext}\Delta h$ where $F_{ext}$ is the magnitude of the external force, the tension in the rope due to pulling on the rope. - $W_g = mg \Delta h$ where $mg$ is the magnitude of the gravitational force and $h$ is the height of the object. - $W_{pulley} = -(mg + F_{ext})\Delta h$ where $m$ is the mass of the box and $F_{ext}$ is the total mass of the pulley and rope. We can see that if the external force is greater than the force of gravity, the work done by the external force is positive. If the external force is less than the force of gravity, the work done by the external force is negative. If the external force is equal to the force of gravity, the work done by the external force is zero. However, this statement is not entirely accurate because the work done by external forces is usually not calculated, since the external force does not include the friction forces acting on both the box and the pulley and rope. The work done by the tension forces within the rope, $F_{ext}$, is also not calculated, since, as we mentioned earlier, we are only considering the external forces acting on the object. **Work Done by a Variable Force** The work done by a variable force $F$ can be calculated by integrating the force over the displacement. This can be written as: $W = \int F(x) dx$ Where $F(x)$ is the force as a function of position $x$. This equation takes into account that the forces acting on the object might be variable, and that they might depend on the position of the object. **Spring Force** A spring is a device that stores elastic potential energy. The force exerted by a spring is always in the opposite direction to the displacement of the spring from its equilibrium position. The magnitude of the spring force is proportional to the displacement, as shown in Figure 5.14. This is called Hooke's Law, and it states that the force of a spring is given by: $F=-kx$ where $k$ is the spring constant and $x$ is the displacement. The negative sign in this equation indicates that the spring force is always opposite in direction to the displacement. **Work Done by the Spring Force** We can calculate the work done by the spring force by integrating over the displacement. The work done by the spring force as the spring is stretched or compressed from an initial position $x_i$ to a final position $x_f$ is: $W_s = \int_{x_i}^{x_f} F(x) dx = \int_{x_i}^{x_f}-kx dx = \frac{1}{2}k(x_i^2 - x_f^2)$ If $x_i$ = 0 and the spring is at its equilibrium position, a positive displacement occurs when $x_f$ > 0 (i.e., the spring is stretched), and the work done by the spring force is negative, $W = -\frac{1}{2}kx_f^2$. If the spring is compressed from its equilibrium position, $x_f$ < 0, and the work done by the spring force is positive, $W = \frac{1}{2}kx_f^2$. In general, if the work done by the spring force is negative, the spring force is doing work on the object to increase the kinetic energy of the object. If the work done by the spring force is positive, the spring force is doing work on the object to decrease the kinetic energy of the object. ### 5.7 Power We can now expand our consideration of energy and work to include a rate, that is, a time derivative of the work. The rate at which work is done is the power. The average power is defined as: $P_{ave} = \frac{\Delta W}{\Delta t}$ where $\Delta W$ is the work done over a time interval $\Delta t$. The instantaneous power is the rate at which work is done at a particular instant in time and is defined as: $P = \lim_{\Delta t \to0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt}$ If $\Delta W$ is the work done by a force acting to an object, the power of the force is the rate at which the the force does work. The units of power are watts (W), where 1 W = 1 J/s. We also use the kilowatt-hour (kWh) as a unit of energy equal to 3.6 x 10^6 J. This is equivalent to a power of 1 kW sustained over a time of 1 h. ### 5.8 Power, Force, and Velocity Power is the rate at which a force does work. Therefore, the power delivered by a force $F$ to an object is the scalar product of the force and the velocity vector: $P = F \cdot v = Fv cos\theta$ where $\theta$ is the wedge between the force vector and the velocity vector. The unit of power is the watt (W), which is defined as 1 J/s.