Mathematics
Work Done by a Constant Force
Work done by a constant force is the product of the force applied and the displacement of the object in the direction of the force. Mathematically, it is represented as the dot product of force and displacement vectors. When the force and displacement are in the same direction, work done is positive; when they are in opposite directions, work done is negative.
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12 Key excerpts on "Work Done by a Constant Force"
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
6.1 Work Done by a Constant Force 159 as the magnitude F of the force times the magnitude s of the displacement: W = Fs. The work done to push a car is the same whether the car is moved north to south or east to west, provided that the amount of force used and the distance moved are the same. Since work does not convey directional information, it is a scalar quantity. The equation W = Fs indicates that the unit of work is the unit of force times the unit of distance, or the newton · meter in SI units. One newton · meter is referred to as a joule (J) (rhymes with “cool”), in honor of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table6.1 summarizes the units for work in several systems of measurement. TABLE 6.1 UnitsofMeasurementforWork System Force × Distance = Work SI newton (N) meter (m) joule (J) CGS dyne (dyn) centimeter (cm) erg BE pound (lb) foot (ft) foot · pound (ft · 1b) The definition of work as W = Fs does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work. Often, the force and displacement do not point in the same direction. For instance, Figure6.2a shows a suitcase-on-wheels being pulled to the right by a force that is applied along the handle. The force is directed at an angle θ relative to the displacement. In such a case, only the component of the force along the displacement is used in defining work.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
Since work does not convey directional information, it is a scalar quantity. The equation W = Fs indicates that the unit of work is the unit of force times the unit of distance, or the newton · meter in SI units. One newton · meter is referred to *When discussing work, it is customary to use the symbol s → for the displacement, rather than x → or y → . F F s FIGURE 6.1 Work is done when a force F → pushes a car through a displacement s → . 6.1 Work Done by a Constant Force 145 as a joule (J) (rhymes with “cool”), in honor of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table 6.1 summarizes the units for work in several systems of measurement. The definition of work as W = Fs does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work. Often, the force and displacement do not point in the same direction. For instance, Figure 6.2a shows a suitcase-on-wheels being pulled to the right by a force that is applied along the handle. The force is directed at an angle relative to the displacement. In such a case, only the component of the force along the displacement is used in defining work. As Figure 6.2b shows, this compo- nent is F cos , and it appears in the general definition below: DEFINITION OF WORK DONE BY A CONSTANT* FORCE The work done on an object by a constant force F → is W = ( F cos θ ) s (6.1) where F is the magnitude of the force, s is the magnitude of the displacement, and is the angle between the force and the displacement. SI Unit of Work: newton · meter = joule (J) When the force points in the same direction as the displacement, then = 0°, and Equation 6.1 reduces to W = Fs.
eBook - PDF- Michael Tammaro(Author)
- 2019(Publication Date)
- Wiley(Publisher)
142 Work and Energy 6 A new kind of farming—wind farming—is gaining popularity as the world turns to renew- able energy sources, due to heightened concerns over the environmental impact and the availability of nonrenewable sources. These wind turbines, part of the Thorntonbank off- shore wind farm located about 28 km off the Belgian coast, will eventually have an energy- producing capacity of 300 megawatts—enough to supply about 150,000 homes with their electricity needs. The concepts of energy, work, and power are explored in detail here in Chapter 6. v.schlichting/Shutterstock Work Done by a Constant Force | 143 6.1 Calculate the Work Done by a Constant Force. When someone is doing work, it usually means that they are engaged in an activity involving a mental or physical labor for the purpose of achieving a result. In physics, forces do work, and work has a very specific meaning that differs from its everyday meaning. To understand the physics definition of work, consider the situation in Animated Figure 6.1.1, where two identical blocks are initially at rest on a horizontal, frictionless surface and are acted upon by a force F . They are prevented from moving by stops. When the animation is run, the stops disappear and the blocks begin to move. Start the animation and watch carefully. 6.1 Work Done by a Constant Force Learning Objective The force F is horizontal in part (a), while it is directed at an angle θ above the horizontal in part (b). The forces act as each block is moved through the same displacement r Δ , as shown. Each force does work on the block as it moves. In the animation, the forces have the same magnitude, but less work is performed in part (b) than in part (a).
eBook - PDF- Richard C. Hill, Kirstie Plantenberg(Authors)
- 2013(Publication Date)
- SDC Publications(Publisher)
Now that we have a physical understanding of what work is, let’s go back and look at Equation 7.1-1 and attempt to gain a mathematical understanding of the work equation. Equation 7.1-1 states that the work U done by an external force F being applied to the particle is equal to the integral of the dot product of F with the differential change in position dr. This most general form of the equation allows for the force to be a function of position and applied at an angle to the displacement. The force F in Equation 7.1-1 may be a single Units of Work SI units: Newton–meter [N-m] Joule [J = 1 N-m] erg [erg = 1 x 10 -7 J] US customary units: foot–pound [ft-lb = 1.3558 J] kilocalorie [kcal = 4187 J] British thermal unit [Btu = 778.16 ft-lb = 1055 J] Conceptual Dynamics Kinetics: Chapter 7 – Particle Work and Energy 7 - 4 force or the summation of several forces. If you don’t recall what a dot product is, refer to Appendix B. One interpretation of the dot product F·dr is that it is the product of the magnitude of dr with the magnitude of the vector F projected onto vector dr. This explanation will make more sense after considering a couple of examples. Note that the result of a dot product of two vectors is a scalar; therefore, the work U done by a force is also a scalar. 7.1.1) Work Done by a Constant Force Equation 7.1-1 gives the most general equation for calculating the work done by a force. To gain a better understanding of how to use this equation, we will examine several special cases. We will begin with an examination of Work Done by a Constant Force F. This force will be constant in both magnitude and direction. This special case will help provide some insight into the notion of work. Since force F is constant, it can be factored out of the integral in Equation 7.1-1 to arrive at Equation 7.1-2. Equation 7.1-2 may be used whenever the force under consideration does not change as a function of position either in direction or magnitude.
eBook - PDF- Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
In this chapter we study the relationship between work and one particular type of energy — the energy of motion of a body, which we call kinetic energy. 11- 2 Work Done by a Constant Force Figure 11-2a shows a block of mass m being lifted through a vertical distance h by a winch that is turned by a motor. The block is raised at a constant velocity; since its acceleration is equal to zero, the net force acting on it is, by Newton’s sec- ond law, also equal to zero. The magnitude of the upward force exerted by the motor and winch must thus equal the magnitude of the downward force m due to gravity. In Fig. 11-2 b, a conveyor belt is operated by a motor to move an identical block a distance L up an incline that makes an angle with the horizontal. If the block moves at a constant velocity, the net force is again zero, and so the magnitude of the force up the incline exerted by the belt must equal the component of the weight mg sin that acts down the incline. In both cases, the final result is the same — the block has been raised a distance h. If we release the block and al- F B g B T B low it to fall, it will reach the ground with a certain speed v. We could use the falling block to accomplish some objec- tive, such as driving a spike into the ground or launching a projectile from a catapult. The outcome would be the same, no matter how the block was originally raised. Once the block has been raised, we can turn the two motors off and the block will remain in place. That is, it costs some fuel or electrical power to run the motors only to lift the block, not to hold it in place. The investment in this process is in the lifting, not in the holding. We define the work W done by a constant force that moves a body through a displacement in the direction of the force as the product of the magnitudes of the force and the displacement: (constant force, (11-1) In Fig. 11-2 a the motor exerts a force of magnitude T mg in moving the block a distance h.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
6.1 | Work Done by a Constant Force 143 As Figure 6.2b shows, this component is F cos u, and it appears in the general definition below: Definition of Work Done by a Constant* Force The work done on an object by a constant force F B is W 5 (F cos u)s (6.1) where F is the magnitude of the force, s is the magnitude of the displacement, and u is the angle between the force and the displacement. SI Unit of Work: newton ? meter 5 joule (J) When the force points in the same direction as the displacement, then u 5 08, and Equa- tion 6.1 reduces to W 5 Fs. Example 1 shows how Equation 6.1 is used to calculate work. Table 6.1 Units of Measurement for Work System Force 3 Distance 5 Work SI newton (N) meter (m) joule (J) CGS dyne (dyn) centimeter (cm) erg BE pound (lb) foot (ft) foot # pound (ft # 1b) *Section 6.9 considers the work done by a variable force. EXAMPLE 1 | Pulling a Suitcase-on-Wheels Find the work done by a 45.0-N force in pulling the suitcase in Figure 6.2a at an angle u 5 50.08 for a distance s 5 75.0 m. Reasoning The pulling force causes the suitcase to move a distance of 75.0 m and does work. However, the force makes an angle of 50.08 with the displacement, and we must take this angle into account by using the definition of work given by Equation 6.1. Solution The work done by the 45.0-N force is W 5 (F cos u)s 5 [(45.0 N)cos 50.0°](75.0 m) 5 2170 J The answer is expressed in newton ? meters or joules (J). The definition of work in Equation 6.1 takes into account only the component of the force in the direction of the displacement. The force component perpendicular to the displacement does no work. To do work, there must be a force and a displacement, and since there is no displacement in the perpendicular direction, there is no work done by the perpendicular component of the force. If the entire force is perpendicular to the dis- placement, the angle u in Equation 6.1 is 908, and the force does no work at all.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
6.1 | Work Done by a Constant Force 127 As Figure 6.2b shows, this component is F cos u, and it appears in the general definition below: Definition of Work Done by a Constant* Force The work done on an object by a constant force F B is W 5 (F cos u)s (6.1) where F is the magnitude of the force, s is the magnitude of the displacement, and u is the angle between the force and the displacement. SI Unit of Work: newton ? meter 5 joule (J) When the force points in the same direction as the displacement, then u 5 08, and Equa- tion 6.1 reduces to W 5 Fs. Example 1 shows how Equation 6.1 is used to calculate work. Table 6.1 Units of Measurement for Work System Force 3 Distance 5 Work SI newton (N) meter (m) joule (J) CGS dyne (dyn) centimeter (cm) erg BE pound (lb) foot (ft) foot # pound (ft # 1b) *Section 6.9 considers the work done by a variable force. EXAMPLE 1 | Pulling a Suitcase-on-Wheels Find the work done by a 45.0-N force in pulling the suitcase in Figure 6.2a at an angle u 5 50.08 for a distance s 5 75.0 m. Reasoning The pulling force causes the suitcase to move a distance of 75.0 m and does work. However, the force makes an angle of 50.08 with the displacement, and we must take this angle into account by using the definition of work given by Equation 6.1. Solution The work done by the 45.0-N force is W 5 (F cos u)s 5 [(45.0 N)cos 50.0°](75.0 m) 5 2170 J The answer is expressed in newton ? meters or joules (J). The definition of work in Equation 6.1 takes into account only the component of the force in the direction of the displacement. The force component perpendicular to the displacement does no work. To do work, there must be a force and a displacement, and since there is no displacement in the perpendicular direction, there is no work done by the perpendicular component of the force. If the entire force is perpendicular to the dis- placement, the angle u in Equation 6.1 is 908, and the force does no work at all.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler, Heath Jones, Matthew Collins, John Daicopoulos, Boris Blankleider(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
* In such a case, the work W is defined as the magnitude F of the force times the magnitude s of the displacement: W = Fs. The work done to push a car is the same whether the car is moved north to south or east to west, provided that the amount of force used and the distance moved are the same. Since work does not convey directional information, it is a scalar quantity. FIGURE 6.1 Work is done when a force F pushes a car through a displacement s. F F s The equation W = Fs indicates that the unit of work is the unit of force times the unit of distance, or the newton ⋅ metre in SI units. One newton ⋅ metre is referred to as a joule (J) (rhymes with ‘cool’), in honour of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table 6.1 summarises the units for work in several systems of measurement. TABLE 6.1 Units of measurement for work System Force × Distance = Work SI newton (N) metre (m) joule (J) CGS dyne (dyn) centimetre (cm) erg BE pound (lb) foot (ft) foot ⋅ pound (ft ⋅ lb) The definition of work as W = Fs does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work. Often, the force and displacement do not point in the same direction. For instance, figure 6.2a shows a suitcase‐on‐wheels being pulled to the right by a force that is applied along the handle. The force is directed at an angle relative to the displacement. In such a case, only the component of the force along the displacement is used in defining work. As figure 6.2b shows, this component is F cos , and it appears in the general definition below.
eBook - PDFEngineering Mechanics
Problems and Solutions
- Arshad Noor Siddiquee, Zahid A. Khan, Pankul Goel(Authors)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
13.1 Introduction In this chapter we will study two principles, i.e., the work and energy principle and the principle of conservation of energy. Earlier, we have used D’ Alembert’s principle to analyse the kinetic problems. The work and energy principle is another alternative method to analyse the kinetic problems. The basic difference in its application is that the work and energy principle is preferred when problems deal especially with velocity. However D’ Alembert’s principle is preferred when problems deal with acceleration. 13.2 Work Done by a Force Work is done by a force when a force displaces a particle by some displacement in its direction. It is a scalar quantity. Consider a particle of mass ‘m’ kg lying on a smooth horizontal surface as shown in Fig. 13.1. If the particle is pushed by a constant force ‘p’ by displacement ‘ d ’ then the work done is given by W = p × d Chapter 13 Work and Energy P P m P mg mg N N d Fig. 13.1 592 Engineering Mechanics However, if the particle is pushed for displacement ‘d’ by a constant force ‘p’ applied at an angle ‘α’ from the horizontal surface as shown in Fig. 13.2, the work done is given by W = Force in the direction of displacement × d W = p cos α × d If α = 0° then it belongs to the earlier case as discussed. If α = 90° then it shows that work done will be equal to zero as force is not causing displacement. For example consider Fig. 13.2, where normal reaction ‘N’ of mass ‘m’ is not producing displacement due to which work done by normal reaction will be zero. Work done will be negative if force and displacement are opposite in direction. For example work done by frictional force is always negative. It is to be noted that force and displacement both are vector quantities but work is a scalar quantity. The unit of work is Nm or Joule in S.I. unit. 13.3 Work Done by a Variable Force Consider a particle displaced by variable force as shown in Fig. 13.3. P m P sin α mg mg N N d α P α P cos α P cos α P sin α Fig.
No longer available |Learn morePhysics for Scientists and Engineers
Foundations and Connections, Extended Version with Modern Physics
- Debora Katz(Author)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
Mathematically, this theorem is written as W tot 5 DK (9.5) Now consider the third scenario, where a constant force is applied to a particle whose displacement is not parallel or antiparallel to the force; our goal is to derive a somewhat more general expression for work. Figure 9.3A shows Paul pulling a sled along the ice. As in the case of the puck, we consider the sled (and its contents) to be a one-particle system and friction between the ice and the sled to be negligible. Paul exerts a constant force F u at an angle u with respect to the x axis (Fig. 9.3B). The sled’s displacement Dx is purely along the x axis (Fig. 9.3C). According to Equation 9.1, we need the x component of Paul’s force, F x 5 F cos u . Then, the work done by him on the sled is W 5 F x Dx 5 1F cos u 2 Dx This example of Paul pulling the sled gives us an idea of how to write a more general expression for the Work Done by a Constant Force F u exerted on a particle whose displacement is Dr u : W 5 F Dr cos u (9.6) where u is the angle between F u and Dr u , and F and Dr are (positive) magnitudes. WORK–KINETIC ENERGY THEOREM ! Underlying Principle FIGURE 9.3 A. Paul pulls a sled. B. The free-body diagram shows that the force exerted by Paul makes an angle u with the x axis. C. The displacement of the sled is along the x axis from x i to x f . A. B. C. y x x x i x f u x u D F u F g u F N u a u We use Dx 5 1x f 2 x i 2 and K 5 1 2 mv 2 (Eqs. 2.1 and 8.1) to rewrite Equation 9.3 in terms of displacement Dx and the change in kinetic energy DK. F P Dx 5 DK (9.4) Equation 9.4 relates the change in the puck’s kinetic energy to the force applied by the player’s stick. In the language of Newton’s second law, we say that the constant force exerted by the stick on the puck accelerated the puck. In the language of the conservation approach, we say that the work done by the stick on the puck changes (increases) the puck’s kinetic energy.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
. . 7.14 Given a variable force as a function of position, cal- culate the work done by it on an object by integrating the function from the initial to the final position of the object, in one or more dimensions. 7.15 Given a graph of force versus position, calculate the work done by graphically integrating from the initial position to the final position of the object. 7.16 Convert a graph of acceleration versus position to a graph of force versus position. 7.17 Apply the work–kinetic energy theorem to situations where an object is moved by a variable force. Learning Objectives ● When the force F → on a particle-like object depends on the position of the object, the work done by F → on the object while the object moves from an initial posi- tion r i with coordinates (x i , y i , z i ) to a final position r f with coordinates (x f , y f , z f ) must be found by integrating the force. If we assume that component F x may depend on x but not on y or z, component F y may depend on y but not on x or z, and component F z may depend on z but not on x or y, then the work is W = ∫ x f x i F x dx + ∫ y f y i F y dy + ∫ z f z i F z dz. ● If F → has only an x component, then this reduces to W = ∫ x f x i F(x)dx. Key Idea Additional examples, video, and practice available at WileyPLUS It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed? KEY IDEAS 1. The work W s done on the canister by the spring force is related to the requested distance d by Eq. 7-26 (W s = − 1 2 kx 2 ), with d replacing x. 2. The work W s is also related to the kinetic energy of the canister by Eq. 7-10 (K f − K i = W ). 3. The canister’s kinetic energy has an initial value of K = 1 2 mv 2 and a value of zero when the canister is momentarily at rest.
eBook - PDF- Stephen Lee(Author)
- 2014(Publication Date)
- CRC Press(Publisher)
● The kinetic energy of a moving object 1 2 mass ( speed ) 2 . Potential energy is the energy which a body possesses because of its position. It may be thought of as stored energy which can be converted into kinetic or other forms of energy. You will meet this again on page 208. The energy of an object is usually changed when it is acted on by a force. When a force is applied to an object which moves in the direction of its line of action, the force is said to do work. For a constant force this is defined as follows. ● The Work Done by a Constant Force force distance moved in the direction of the force . The following examples illustrate how to use these ideas. A brick, initially at rest, is raised by a force averaging 40 N to a height 5 m above the ground where it is left stationary. How much work is done by the force? S OLUTION The work done by the force raising the brick is 40 5 200 J. Figure 9.1 Examples 9.2 and 9.3 show how the work done by a force can be related to the change in kinetic energy of an object. A train travelling on level ground is subject to a resisting force (from the brakes and air resistance) of 250 kN for a distance of 5 km. How much kinetic energy does the train lose? 5 m 40 N 202 AN INTRODUCTION TO MATHEMATICS FOR ENGINEERS : MECHANICS E XAMPLE 9.1 E XAMPLE 9.2 S OLUTION The forward force is 250 000 N. The work done by it is 250 000 5000 1 250 000 000 J. Hence 1 250 000 000 J of kinetic energy are gained by the train, in other words 1 250 000 000 J of kinetic energy are lost and the train slows down. This energy is converted to other forms such as heat and perhaps a little sound. A car of mass m kg is travelling at u ms 1 when the driver applies a constant driving force of F N. The ground is level and the road is straight and air resistance can be ignored. The speed of the car increases to v ms 1 in a period of t s over a distance of s m.
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