### SAT Physics Energy, Work, and Power - Mechanical Energy

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**SAT Physics Energy, Work, and Power - Mechanical Energy**

**MECHANICAL ENERGY**

Several forms of energy are addressed in a first-year physics course. These include the different types of mechanical energy, electrical energy, thermal energy, heat, light energy, and nuclear energy. At this point, only the different types of mechanical energy will be discussed. The other forms of energy will be explored in subsequent chapters. Mechanical energy is the sum of the kinetic and potential energy of a system. The system is simply the object, or mass, under investigation.

**Kinetic Energy**

Kinetic energy, K, is the energy possessed by moving objects. If an object with mass m is moving at a speed v, its kinetic energy is

^{2}. Kinetic energy is a scalar. Therefore, the direction of an object’s velocity is not important. Only the magnitude of velocity (speed) is significant. In addition, kinetic energy is always positive.

**Potential Energy**

**Potential energy**, U, is the energy possessed by an object based on its position. To possess useful potential energy, an object must be in a position where it will move when released from rest. When the object is released, it gains kinetic energy at the expense of potential energy. If an object has the potential to create kinetic energy, the object has potential energy. There are two mechanical forms of potential energy featured on the SAT Subject Test in Physics: gravitational potential energy and elastic potential energy.

**Gravitational Potential Energy**

**Gravitational potential energy**, U

_{g}, is due to the position of an object in a gravity field. A mass, m, in Earth’s gravity field, g, is pulled toward Earth by the force of gravity, F

_{g}. When the mass is raised to a height of h, the mass is in a position where it will accelerate toward Earth if released. At a height of h, the mass has the potential to create kinetic energy. If the mass is moved to a greater height, the mass will have a greater potential to increase both the final speed and the final kinetic energy. Height is the key factor determining the magnitude of gravitational potential energy.

U

Gravitational potential energy is a scalar. Only the height is important. Technically, both height and gravitational potential energy can have negative values. This depends on the location defined as zero height. To avoid working with negative heights, set the zero point for height measurements at the lowest point that the object reaches during a problem._{g}= mgh**Elastic Potential Energy**

Elastic potential energy, U

_{s}, is the potential energy associated with elastic devices, such as springs. If a spring is stretched or compressed a distance of x, a restoring force, F

_{s}, is generated in the spring according to Hooke’s law.

^{2}. Elastic potential energy is a positive scalar. The direction the spring is displaced (stretched or compressed) does not matter.

**Total Mechanical Energy**

Total mechanical energy, is the sum of the kinetic and potential energies in a system.Total mechanical energy is an instantaneous value. Freeze the action in a problem and assess if the object you are interested in has height (h), speed (v), and/or is attached to a spring displaced a distance (x) . Next, sum the energy equations corresponding to the key variables that appear in the problem. This is most often done at the start and/or end of a problem.

E = K + U

Regardless of its form, energy is expressed in units of joules (N • m). As will soon be demonstrated, mechanical energy can change forms through a process known as work. Energy is known as a state function. Work is a process through which the state of energy may be changed.**Recognizing and Calculating Mechanical Energy**

**WHAT'S THE TRICK?**

Look for height (h), speed (v), and spring displacement (x). If any of these are present, include the corresponding energy in the answer.

**Point A :**The pendulum has height above the lowest point in the swing but it is at rest. Only gravitational potential energy is present. The total mechanical energy at point A is

E = U

_{g}= mgh**Point B :**The pendulum has both speed and height. Both kinetic energy and gravitational potential energy are present. The total mechanical energy at point B is

**Point C:**The pendulum has lost all its height. Only speed remains, so only kinetic energy is present. The total mechanical energy at point C is

^{}

**Assessing the Effect of Changing Values on Energy**

**WHAT'S THE TRICK?**

The key is assessing the relationships in the formula.

^{2}.

**WORK**

When a force is applied to an object, the force can accelerate the object, changing the object’s speed and its kinetic energy. A force can also lift an object to a height h, or it can displace a spring through a distance x. In all these cases, a force displaces an object and changes the object’s total energy.

**Work**is the process of applying a force through a distance to change the energy of an object. Positive work is associated with an increase in the speed of an object. Negative work is associated with a decrease in speed. You can solve for work in two principal ways. The first method involves force and displacement. The second focuses on the change in energy of the object.

Work, W, is the product of the average force, F

_{avg}, applied to an object and the component of displacement parallel to the average force, d

_{parallel}.

W = F

_{avg}d_{parallel}- In most problems, the given force will be equal to the average force needed in the formula. The single exception in introductory physics courses is the restoring force of springs.
- Work requires a change in position known as displacement, which may be indicated in several ways (d, x, or h). It is important to recognize that work involves motion and requires objects to be displaced.
- Both force and displacement are vectors, and vectors at angles can be split into component vectors. The most important aspect of solving for work is remembering to
**use parallel force and displacement vectors**or the parallel components of these vectors. - Forces do work only when they are parallel to the displacement of the object. Forces perpendicular to displacement result in no work.
- Work is positive when the components of force and displacement both point in the same direction. Work is negative when the components of these vectors point in opposite directions.

**Solving for Work Using Force and Displacement**

In the figure above, mass m is being pulled horizontally to the right by force F, inclined at angle θ. Force F and its components are shown in the diagram. The surface is rough. Friction, f, opposes the motion.

**(A)**Determine the work done by the applied force, W

_{F}, as it moves the mass m through displacement d.

**WHAT'S THE TRICK?**

The applied force, F, is not parallel to the displacement. However, the x-component of the applied force, F cos θ, is parallel to the displacement and has the same direction as the displacement. F cos θ is capable of doing positive work, increasing the speed and kinetic energy of the object.

W

_{F}= ( + F cos θ)d**(B)**Determine the work of friction, W

_{f}.

**WHAT'S THE TRICK?**

The friction vector, f, is opposite the displacement. It opposes motion, slows the object, decreases the object’s kinetic energy, and performs negative work.

W

_{f}= (-f)d**(C)**Determine the net work, W

_{net}.

**WHAT'S THE TRICK?**

The net work is the total work done on the object. One way to solve for the net work is to use the net force (sum of parallel forces, ) acting on the object.

All other forces (force of gravity, normal force, and F sin θ) are perpendicular to motion. Perpendicular forces and perpendicular components of force do no work.

**Work of Gravity**

Gravity near the surface of Earth is a good example of the

**work done by a constant force**. Earth’s gravity, g, is essentially constant for small changes in height. When a force is constant, the average force equals the value of the constant force,

F

_{avg}= F

_{g}= mg. Displacement parallel to the force of gravity is equal to the change in height, d=Δh. Changes in height are also associated with changes in gravitational potential energy. The work done by gravity can be solved either as a force through a distance or as a change in gravitational potential energy.

W= F

The work done by gravity depends on only a change in height, Δh. As a result, the work done by gravity for horizontal motion is always zero. When objects follow any path consisting of a combination of vertical and horizontal motion, the work done by gravity depends on only the vertical change in height. Although there is a negative sign in the formula, the resulting sign on work also depends on the sign of the change in height, Δh = h_{avg}d_{parallel}W = -ΔU_{g }W_{g}= -mgΔh W_{g}= -mgΔh_{f }- h

_{i}Ultimately, the work done by gravity is positive for downward motion, when force and vertical displacement have the same direction. It is negative for upward motion, when force and vertical displacement oppose each other.

Most questions regarding vertical motion involve the

**work done by an external applied force that acts to lift an object.**When an object is lifted, the work done by an external applied force, F, is done against gravity. If the object starts at rest and finishes at rest or if the object moves at constant velocity, the work done by the external applied force will have the same magnitude as the work done by gravity. However, work will have the opposite sign.

W

The work done by an external force to lift an object is positive since force and displacement have the same direction._{F }= mg Δh**The Work Of Gravity**

**WHAT'S THE TRICK?**

Two displacements are occurring, Δh and d. Use the change in height,Δh,since it is parallel to the force of gravity.

W = F

W

The sign on work is positive in this case. Why? You can determine the sign on work using two methods._{avg}d_{parallel }W = f_{g}ΔhW

_{g}= mgΔh**Vector method:**If the force and the displacement vectors point in the same direction, work is positive. When they are opposite each other, work is negative. F_{g}and Ah have the same direction. So the work is positive.**Energy method:**If the kinetic energy increases during a displacement, the work is positive. Gravity increases the speed of the mass, so the work of gravity is positive.

**Work of a Spring**

The work of a spring, W

_{s}, is done when a spring is stretched or compressed through a displacement x, thereby changing its length. The work of a spring is a good example of the

**work done by a variable force**. Springs obey Hooke’s law, F

_{s}= kx. The restoring force, F

_{s}, is the instantaneous force in a spring at a specific spring length, x. Changing spring length, Δx, changes the restoring force as shown in Figure 7.1.

**Figure 7.1.**Restoring Force vs. Displacement

_{avg}. The average force is simply half of the force needed to change the length of the spring.

Questions may instead focus on the

**work done on the spring**by an external applied force, F, which acts to stretch or compress the spring from its equilibrium position. The external applied force that stretches or compresses a spring must be equal in magnitude, but opposite in direction, to the resulting restoring force created as the length of the spring changes. As a result, the work done on a spring by an external applied force is equal to the work done by the spring when it is released and moves to restore to equilibrium.

**The Work of a Spring**

**WHAT'S THE TRICK?**

This problem involves a spring undergoing a displacement (stretch or compression). Work always involves change. The work done to a spring depends on the change in length of the spring,

Δx = x

_{f }- x

_{i}

_{i}= 0.

**Interpreting a Force vs. Displacement Graph**

Recall from Chapter 1 that slopes of lines and areas under curves have significance. The slope of a force vs. displacement graph (Figure 7.1) has units of N/m. These are the same units as the spring constant (k). The area underneath Figure 7.1 has units of N • m, or joules. These are the same units as for work and energy.

**Figure 7.2**Force-displacement graph

W = F

_{avg }d = height × base**Work and Force-Displacement Graphs**

The force and displacement are graphed above. Determine the work done on the mass by the variable force.

**WHAT'S THE TRICK?**

Determine the area of the triangle formed by the graph.

The one-half in the formula actually solves for the average force.

**Work-Kinetic Energy Theorem**

The

**work-kinetic energy theorem**states that work is equal to the change in kinetic energy. When one or more unbalanced forces act on a mass, the mass will accelerate, changing in both speed and kinetic energy. Several forms of work may be present (W

_{p}, W

_{g}, W

_{s}, and W

_{f}) and can be calculated. The net work, W

_{net}, done on the mass will be equal to the change in kinetic energy, ΔK, resulting from the total work.

W

Using the work-kinetic energy theorem is a quick way to find the net work done on a mass when the initial and final speeds are known. It can also be used to find the final speed of a mass if the net work and initial velocity are known. Since the net work is tied to changes in kinetic energy and changes in speed, a mass must accelerate in order for net work to be non-zero. Thus, when an object moves at constant velocity, the net work is always equal to zero._{net}= ΔK**Changes in Speed and Net Work (Total Work)**

**WHAT'S THE TRICK?**

An initial speed is given, and a final speed is requested. This implies a change in speed. There are two ways to solve for final speed. One involves force and kinematics. The other involves work and energy. The key in this problem is the 20 joules of work. No value for force is given. Try using the work-kinetic energy theorem:

**Work in Uniform Circular Motion**

When a mass follows a curved path, such as a circle, the instantaneous velocity and displacement vectors are directed tangent to the curved path. The net force is centripetal and is directed toward the center of the circle. As result, the net (centripetal) force and the displacement are always perpendicular to each other, as shown in Figure 7.2.

**Figure 7.3.**Vectors in uniform, circular motion

W

_{net}= ΔK, results in the same conclusion. In uniform circular motion speed is constant. If speed is constant, then the change in kinetic energy and the net work are both zero.

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