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As he got older, he decided to build and use a staircase instead. (d) Can static friction produce positive work? In this circumstance, the work to stretch the spring from 0 to 12 cm is also equal to the work for a composite path from 0 to 6 cm followed by an additional stretch from 6 cm to 12 cm. 7. So why do you eventually feel tired just holding the briefcase, if youre not doing any work on it? For an object moving on a surface, the displacement [latex]d\mathbf{\overset{\to }{r}}[/latex] is tangent to the surface. In many cases, it is convenient to express the dot product for gravitational work in terms of the x-, y-, and z-components of the vectors. A particle moves along a curved path [latex]y(x)=(10\,\text{m})\left\{1+\text{cos}[(0.1{\,\text{m}}^{-1})x]\right\},[/latex] from [latex]x=0[/latex] to [latex]x=10\pi \,\text{m,}[/latex] subject to a tangential force of variable magnitude [latex]F(x)=(10\,\text{N})\text{sin}[(0.1{\,\text{m}}^{-1})x]. We can use the equation of the path to express y and dy in terms of x and dx; namely. which the figure also illustrates as the horizontal component of the force times the magnitude of the displacement. The spring exerts a force in the opposite direction to (b) an extension or stretch, and (c) a compression. The quantity 1 2mv2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m moving at a speed v. ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) [/latex], [latex]W=\frac{1}{2}k({x}_{B}^{2}-{x}_{A}^{2}). That is, the actual path could involve going back and forth before ending. citation tool such as, Authors: William Moebs, Samuel J. Ling, Jeff Sanny. We can evaluate the difference in height to answer (a) and (b). The infinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacement along the path. A constant 20-N force pushes a small ball in the direction of the force over a distance of 5.0 m. What is the work done by the force? Thus the Kinetic energy of . At what point(s) does the force of gravity do positive work on the mass? (a) The spring exerts no force at its equilibrium position. By Manish Naik Work done by friction comprises the moving body's displacement opposite to the direction of friction force. (Then positive x corresponds to a stretch and negative x to a compression.) How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement of the object, relative to the floor, and do positive work. Equation \ref{7.2} defines the total work as a line integral, or the limit of a sum of infinitesimal amounts of work. The infinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacement along the path, \[dW = F_{x} dx + F_{y} dy + F_{z} dz \ldotp \nonumber\]. Which choice is more convenient depends on the situation. Creative Commons Attribution License This quantity is our first example of a form of energy. is 0.220. You lift an oversized library book, weighing 20 N, 1 m vertically down from a shelf, and carry it 3 m horizontally to a table (Figure \(\PageIndex{4}\)). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. A typical coordinate system has the x-axis horizontal and the y-axis vertically up. With this choice of coordinates, the spring force has only an x-component, Fx = kx, and the work done when x changes from xA to xB is, \[W_{spring,\; AB} = \int_{A}^{B} F_{x} dx = - k \int_{A}^{B} xdx = -k \frac{x^{2}}{2} \Big|_{A}^{B} = - \frac{1}{2} k \big( x_{B}^{2} - x_{A}^{2} \big) \ldotp \label{7.5}\]. Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion. Work is done when a force acts on something that undergoes a displacement from one position to another. and you must attribute OpenStax. Static friction is the force of friction on an object that is not moving. With this choice of coordinates, the spring force has only an x-component,[latex]{F}_{x}=\text{}kx[/latex], and the work done when x changes from [latex]{x}_{A}[/latex] to [latex]{x}_{B}[/latex] is. The reason is that forces like friction are classified as nonconservative forces, or dissipative forces, as we discuss in the next chapter. 7.1 Work Copyright 2016 by OpenStax. For part (a), [latex]{x}_{A}=0[/latex] and [latex]{x}_{B}=6\text{cm}[/latex]; for part (b), [latex]{x}_{B}=6\text{cm}[/latex] and [latex]{x}_{B}=12\text{cm}[/latex]. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . What was the total work done against friction moving the couch away from its original position and back again? Can kinetic friction ever be a constant force for all paths? A particle moving in the xy-plane is subject to a force, [latex]\mathbf{\overset{\to }{F}}(x,y)=(50\,\text{N}\cdot {\text{m}}^{2})\frac{(x\mathbf{\hat{i}}+y\mathbf{\hat{j}})}{{({x}^{2}+{y}^{2})}^{3\text{/}2}},[/latex]. (Hint: Consult a table of integrals or use a numerical integration program.). (Lift is used as opposed to drop.). At what point(s) does the force of gravity do What is the work done by this force? (b) Can kinetic friction do positive work? Energy is consumed, but no energy is transferred. Calculate the work done on the particle by this force, as it moves in a straight line from the point (3 m, 4 m) to the point (8 m, 6 m). We first define the increment of work dW done by a force [latex]\mathbf{\overset{\to }{F}}[/latex] acting through an infinitesimal displacement [latex]d\mathbf{\overset{\to }{r}}[/latex] as the dot product of these two vectors: Then, we can add up the contributions for infinitesimal displacements, along a path between two positions, to get the total work. Explain. The part of the contact force on the object that is perpendicular to the surface is the normal force [latex]\mathbf{\overset{\to }{N}}. Figure 7.3(b) shows a person holding a briefcase. If youre driving your car at the speed limit on a straight, level stretch of highway, the negative work done by kinetic friction of air resistance is balanced by the positive work done by the static friction of the road on the drive wheels. (a) How much work is done by friction as the sled moves 30.0 m along the hill? In this case, we can factor out the force; the remaining integral is just the total displacement, which only depends on the end points A and B, but not on the path between them: We can also see this by writing out Figure in Cartesian coordinates and using the fact that the components of the force are constant: Figure(a) shows a person exerting a constant force [latex]\mathbf{\overset{\to }{F}}[/latex] along the handle of a lawn mower, which makes an angle [latex]\theta[/latex] with the horizontal. Work-Energy Theorem. As a young man, Tarzan climbed up a vine to reach his tree house. A typical coordinate system has the x-axis horizontal and the y-axis vertically up. then you take the work due to friction so 60N*500m and get 30000J. However, the total work was not zero. No, its only approximately constant near Earths surface. However, if the object is more than a particle, and has an internal structure, the normal contact force can do work on it, for example, by displacing it or deforming its shape. Wnet = 1 2mv2 1 2mv2 0. (Although we chose to illustrate dW in Cartesian coordinates, other coordinates are better suited to some situations.) [/latex], [latex]W=0.54\,\text{J}=\frac{1}{2}k[{(6\,\text{cm})}^{2}-0][/latex], so [latex]k=3\,\text{N/cm}\text{. a. The components of the force are given functions of x and y. For part (a), xA=0xA=0 and xB=6cmxB=6cm; for part (b), xB=6cmxB=6cm and xB=12cmxB=12cm. The part of the contact force on the object that is perpendicular to the surface is the normal force N.N. Forces like gravity (those that do zero work over any closed path) are classified as conservative forces and play an important role in physics. The person must exert an upward force, equal in magnitude to the weight of the briefcase, but this force does no work, because the displacement over which it acts is zero. The components of the force are given functions of x and y. We can also see this by writing out Equation \ref{7.2} in Cartesian coordinates and using the fact that the components of the force are constant: \[\begin{split} W_{AB} & = \int_{path\; AB} \vec{F} \cdotp d \vec{r} = \int_{path\; AB} (F_{x} dx + F_{y} dy + F_{z} dz) = F_{x} \int_{A}^{B} dx + F_{y} \int_{A}^{B} dy + F_{z} \int_{A}^{B} dz \\ & = F_{x} (x_{B} - x_{A}) + F_{y} (y_{B} - y_{A}) + F_{z} (z_{B} - z_{A}) = \vec{F} \cdotp (\vec{r}_{B} - \vec{r}_{A}) \ldotp \end{split} \nonumber\], Figure \(\PageIndex{2a}\) shows a person exerting a constant force \(\vec{F}\) along the handle of a lawn mower, which makes an angle \(\theta\) with the horizontal. Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. Show transcribed image text Expert Answer 100% (1 rating) Top Expert 500+ questions answered This will be mentioned in the next chapter. 1 Answer +1 vote answered Mar 14, 2019 by Devanshi (68.0k points) selected Mar 17, 2019 by faiz Best answer Generally work done by the kinetic friction on an object is negative because the displacement is always opposite the friction force. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work. However, in many other sources, the work done by friction can be calculated. As an external force, static friction can do work. As expected, this is exactly the same result as before. The physical concept of work is straightforward: you calculate the work for tiny displacements and add them up. The force, angle, and displacement are given, so that only the work W is unknown. The bottom line is that you need to analyze each particular case to determine the work done by the forces, whether positive, negative or zero. [/latex], [latex]W=\text{}(20\,\text{N})(-1\,\text{m})=20\,\text{J}\text{. My book says that "In fact, the displacement of the point of application of the friction force is not calculable and so neither is the work done by the friction force.". We could equally well have expressed the dot product in terms of the various components introduced in Vectors. The spring force is the opposite direction to a compression (as it is for an extension), so the work it does is negative. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The normal force and . Explain why it does no work. The units of work are units of force multiplied by units of length, which in the SI system is newtons times meters, [latex]\text{N}\cdot \text{m.}[/latex] This combination is called a joule, for historical reasons that we will mention later, and is abbreviated as J. How much work is done against the gravitational force on a 5.0-kg briefcase when it is carried from the ground floor to the roof of the Empire State Building, a vertical climb of 380 m? Answers and Replies Oct 7, 2005 #2 In this example, everything was given in terms of x and y-components, which are easiest to use in evaluating the work in this case. where |lAB| is the path length on the surface. We have just seen that the work done by a constant force of gravity depends only on the weight of the object moved and the difference in height for the path taken, WAB = mg(yB yA). Then, the integral for the work is just a definite integral of a function of x. (Lift is used as opposed to drop.). In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement of the object, relative to the floor, and do positive work. Unlike friction or other dissipative forces, described in Example \(\PageIndex{2}\), the total work done against gravity, over any closed path, is zero. In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. [/latex] (Then positive x corresponds to a stretch and negative x to a compression.) Forces between molecules, or in any system undergoing small displacements from a stable equilibrium, behave approximately like a spring force. Gravity does negative work on an object that moves upward ([latex]{y}_{B} \gt {y}_{A}[/latex]), or, in other words, you must do positive work against gravity to lift an object upward. There is zero difference in height for any path that begins and ends at the same place on the shelf, so W = 0. - Reimagining Education Is friction positive or negative work? (Note that if the displacement drdr did have a relative component perpendicular to the surface, the object would either leave the surface or break through it, and there would no longer be any normal contact force. In two dimensions, these were the x- and y-components in Cartesian coordinates, or the r- and \(\varphi\)-components in polar coordinates; in three dimensions, it was just x-, y-, and z-components. Positive work is done against gravity on the upward parts of a closed path, but an equal amount of negative work is done against gravity on the downward parts. This will be mentioned in the next chapter. (b) How much work is required to stretch it an additional 6 cm? The spring in Example \(\PageIndex{5}\) is compressed 6 cm from its equilibrium length. The magnitude of one of these areas is just one-half the triangles base, along the x-axis, times the triangles height, along the force axis. The net work on a system equals the change in the quantity 1 2mv2. If it takes 22.0 kJ of work to stretch the cord by 16.7 m, determine the value of the constant a. However, the total work was not zero. }[/latex], [latex]{W}_{\text{spring},AB}={\int }_{A}^{B}{F}_{x}dx=-k{\int }_{A}^{B}xdx=\text{}k{\frac{{x}^{2}}{2}|}_{A}^{B}=-\frac{1}{2}k({x}_{B}^{2}-{x}_{A}^{2}). In equation form, the translational kinetic energy, KE = 1 2mv2, 7.12. Recall that, in general, a one-dimensional integral is the limit of the sum of infinitesimals, f(x)dx , representing the area of strips, as shown in Figure \(\PageIndex{7}\). This will be mentioned in the next chapter. (c) What is the work done by the gravitational force on the sled? Notice that the work required to stretch the spring from 0 to 12 cm is four times that required to stretch it from 0 to 6 cm, because that work depends on the square of the amount of stretch from equilibrium, \(\frac{1}{2}\)kx2. Gravity does positive work (20 J) when the book moves down from the shelf. The person must exert an upward force, equal in magnitude to the weight of the briefcase, but this force does no work, because the displacement over which it acts is zero. The answer is that muscle fibers in your arm are contracting and doing work inside your arm, even though the force your muscles exert externally on the briefcase doesnt do any work on it. [/latex], [latex]W=(75.0\,\text{N})(25.0\,\text{m})\text{cos}(35.0^\circ)=1.54\times {10}^{3}\,\text{J}\text{. As expected, this is exactly the same result as before. The force of friction opposes the displacement (always), and does negative work. The simplest work to evaluate is that done by a force that is constant in magnitude and direction. This book uses the 700 J; b. For this object sliding along the surface, kinetic friction \(\vec{f}_{k}\) is opposite to d \(\vec{r}\), relative to the surface, so the work done by kinetic friction is negative. When you walk, the force of static friction exerted by the ground on your back foot accelerates you for part of each step. The bottom line is that you need to analyze each particular case to determine the work done by the forces, whether positive, negative or zero. Once your push exceeds the maximum possible static friction (budging force = N), then the block will start moving. As a result, the work done by a force can be positive, negative, or zero, depending on whether the force is generally in the direction of the displacement, generally opposite to the displacement, or perpendicular to the displacement. Gravity does zero work (0 J) when the book moves horizontally from the shelf to the table and negative work (20 J) when the book moves from the table back to the shelf. (e) What is the total work done on the cart? (a) A pendulum swings from left to right, as shown below. Strategy and Concept for Part 2 The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. Kinetic energy is not a vector. consent of Rice University. ), The part of the contact force on the object that is parallel to the surface is friction, f.f. The work done against a force may also be viewed as the work required to overcome this force, as in How much work is required to overcome?) The force of static friction, however, can do positive or negative work. (a) What is the magnitude of the force exerted to keep the car moving at constant speed? Notice that WABWAB depends only on the starting and ending points, A and B, and is independent of the actual path between them, as long as it starts at A and ends at B. From the properties of vectors, it doesnt matter if you take the component of the force parallel to the displacement or the component of the displacement parallel to the forceyou get the same result either way. In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. In part (a), the work is given and you can solve for the spring constant; in part (b), you can use the value of k, from part (a), to solve for the work. Which choice is more convenient depends on the situation. If you push on a stationary block and it doesn't move, it is being held by static friction which is equal and opposite to your push. Is the work done by friction in pure rolling always zero? Since the cosine of the angle between the normal and the tangent to a surface is zero, we have, The normal force never does work under these circumstances. Ask Question Asked 8 years, 1 month ago Modified 8 years, 1 month ago Viewed 11k times 3 By definition, the work done by a force is W = F d, so how can static friction do work? Describe a situation in which a force is exerted for a long time but does no work. Gravity does negative work on an object that moves upward (yB>yAyB>yA), or, in other words, you must do positive work against gravity to lift an object upward. All Rights Reserved. Give an example of a situation in which there is a force and a displacement, but the force does no work. The work done by a force is the integral of the force with respect to displacement along the path of the displacement: \[W_{AB} = \int_{path\; AB} \vec{F} \cdotp d \vec{r} \ldotp \label{7.2}\]. Sometimes the mathematics can seem complicated, but the following example demonstrates how cleanly they can operate. Near the surface of Earth, the gravitational force on an object of mass m has a constant magnitude, mg, and constant direction, vertically down. The coefficient of kinetic friction between the box and surface is [latex]{\mu }_{K}=0.50. No, only its magnitude can be constant; its direction must change, to be always opposite the relative displacement along the surface. If youre driving your car at the speed limit on a straight, level stretch of highway, the negative work done by kinetic friction of air resistance is balanced by the positive work done by the static friction of the road on the drive wheels. In other situations, magnitudes and angles might be easier. (Then positive x corresponds to a stretch and negative x to a compression.) Positive work - The work done by static friction on an object moving along with (without slipping) an accelerating belt is positive with respect to the ground frame as the displacement and the friction force act along the same direction (along the direction of acceleration). To calculate the work done by a spring force, we can choose the x-axis along the length of the spring, in the direction of increasing length, as in Figure, with the origin at the equilibrium position [latex]{x}_{\text{eq}}=0. Then, the integral for the work is just a definite integral of a function of x. We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle between them, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes and angles. (b) When youre finished, you move the book in a straight line back to its original place on the shelf. * (a) Work done by kinetic friction is often negative. Both require the same gravitational work, but the stairs allow Tarzan to take this work over a longer time interval and hence gradually exert his energy, rather than dramatically by climbing a vine. (b) You dont like the new position, so you move the couch straight back to its original position (B to A in Figure). Positive work is done against gravity on the upward parts of a closed path, but an equal amount of negative work is done against gravity on the downward parts. Is energy transferred or changed in form in your example? The other force on the lawn mower mentioned above was Earths gravitational force, or the weight of the mower. Therefore, the work done by gravity on an object is the dot product of its weight and its displacement. -What is the work done by friction as the block slides down the ramp? As an Amazon Associate we earn from qualifying purchases. The vectors involved in the definition of the work done by a force acting on a particle are illustrated in Figure \(\PageIndex{1}\). The segments of the paths are the sides of a right triangle, so the path lengths are easily calculated. The force of static friction does no work in the reference frame between two surfaces because there is never displacement between the surfaces. How can static friction do work? Be certain to include the work he does on the crate and on his body to get up the ramp. However, if the object is more than a particle, and has an internal structure, the normal contact force can do work on it, for example, by displacing it or deforming its shape. The simplest work to evaluate is that done by a force that is constant in magnitude and direction. Explain. Find the work done to accomplish this task. In order to . In Equation 7.5, since F=kxF=kx is a straight line with slope kk, when plotted versus x, the area under the line is just an algebraic combination of triangular areas, where areas above the x-axis are positive and those below are negative, as shown in Figure 7.9. Zero work is done when the displacement of the body in the direction of the force is zero. }[/latex], [latex]\begin{array}{cc}\hfill {W}_{AB}& =\underset{\text{path}\,AB}{\int }\mathbf{\overset{\to }{F}}\cdot d\mathbf{\overset{\to }{r}}=\underset{\text{path}\,AB}{\int }({F}_{x}dx+{F}_{y}dy+{F}_{z}dz)={F}_{x}{\int }_{A}^{B}dx+{F}_{y}{\int }_{A}^{B}dy+{F}_{z}{\int }_{A}^{B}dz\hfill \\ & ={F}_{x}({x}_{B}-{x}_{A})+{F}_{y}({y}_{B}-{y}_{A})+{F}_{z}({z}_{B}-{z}_{A})=\mathbf{\overset{\to }{F}}\cdot ({\mathbf{\overset{\to }{r}}}_{B}-{\mathbf{\overset{\to }{r}}}_{A}).\hfill \end{array}[/latex], [latex]W=Fd\,\text{cos}\,\theta . }[/latex], [latex]d{W}_{\text{N}}=\mathbf{\overset{\to }{N}}\cdot d\mathbf{\overset{\to }{r}}=\mathbf{\overset{\to }{0}}. Sometimes the mathematics can seem complicated, but the following example demonstrates how cleanly they can operate. This integral was not hard to do. We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle between them, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes and angles. (Note that, especially if the work done by a force is negative, people may refer to the work done against this force, where [latex]d{W}_{\text{against}}=\text{}d{W}_{\text{by}}[/latex]. Possibility 1: Yes, because else what is it then that moves your car forward when you accelerates? Usually, there are several ways to do this, which may be more or less convenient, depending on the particular case. Assume no friction acts on the wagon. The spring in Example 7.5 is compressed 6 cm from its equilibrium length. For an object moving on a surface, the displacement d \(\vec{r}\) is tangent to the surface. (There are quotation marks around area because this base-height product has the units of work, rather than square meters.). Alternately, gravity does positive work on an object that moves downward (yB < yA), or you do negative work against gravity to lift an object downward, controlling its descent so it doesnt drop to the ground. }[/latex] (Recall that the spring constant is the slope of the force F(x) versus its stretch x.) We can evaluate the difference in height to answer (a) and (b). So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m/s. Positive work is done by a force when the displacement of the body is in the direction of the force applied whereas negative work is done by the force when the direction of the displacement of the body is opposite to the direction of the force. The work done on the lawn mower is, \[W = \vec{F} \cdotp \vec{d} = Fd \cos \theta,\nonumber \]. If you continue to push on a wall without breaking through the wall, you continue to exert a force with no displacement, so no work is done. The force, angle, and displacement are given, so that only the work W is unknown. Notice that [latex]{W}_{AB}[/latex] depends only on the starting and ending points, A and B, and is independent of the actual path between them, as long as it starts at A and ends at B. Forces can vary as a function of position, and displacements can be along various paths between two points. f s (max) = s N = ( 0.45) ( 980 N) = 440 N. to move the crate. (There are quotation marks around area because this base-height product has the units of work, rather than square meters.). (The energy content of gasoline is about 140 MJ/gal.) A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. (c) What is the work done on the cart by the shopper? Suppose that the sled plus passenger of the preceding problem is pushed 20 m across the snow at constant velocity by a force directed [latex]30^\circ[/latex] below the horizontal. Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of [latex]20.0^\circ[/latex] with the horizontal (see below). The mass is 90kg so divide both sides by 90 and get v^2=187.8889. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work. He pushes in a direction [latex]25.0^\circ[/latex] below the horizontal. One very important and widely applicable variable force is the force exerted by a perfectly elastic spring, which satisfies Hookes law [latex]\mathbf{\overset{\to }{F}}=\text{}k\Delta \mathbf{\overset{\to }{x}},[/latex] where k is the spring constant, and [latex]\Delta \mathbf{\overset{\to }{x}}=\mathbf{\overset{\to }{x}}-{\mathbf{\overset{\to }{x}}}_{\text{eq}}[/latex] is the displacement from the springs unstretched (equilibrium) position (Newtons Laws of Motion). If so, give an example. Solving for x and dx, in terms of y, along the parabolic path, we get, The components of the force, in terms of y, are, so the infinitesimal work element becomes, The integral of [latex]{y}^{1\text{/}2}[/latex] is [latex]\frac{2}{3}{y}^{3\text{/}2}[/latex], so the work done from A to B is. The part of the contact force on the object that is perpendicular to the surface is the normal force \(\vec{N}\). (b) How much work is required to stretch it an additional 6 cm? Work required means work done against the spring force, which is the negative of the work in Equation \ref{7.5}, that is, \[W = \frac{1}{2} k (x_{B}^{2} - x_{A}^{2}) \ldotp \nonumber\]. (a) You first push the couch 3 m parallel to a wall and then 1 m perpendicular to the wall (A to B in Figure \(\PageIndex{3}\)). Then, we can add up the contributions for infinitesimal displacements, along a path between two positions, to get the total work. 1999-2023, Rice University. The net work is the total of my positive work and fiction's negative work, i.e. Gravity does negative work on an object that moves upward (yB > yA), or, in other words, you must do positive work against gravity to lift an object upward. Lets consider the work done by these forces in general. Highlights Learning Objectives By the end of this section, you will be able to: Describe the general characteristics of friction List the various types of friction Calculate the magnitude of static and kinetic friction, and use these in problems involving Newton's laws of motion Gravity does zero work (0 J) when the book moves horizontally from the shelf to the table and negative work (20 J) when the book moves from the table back to the shelf. How much work is done on the lawn mower by the person in Figure \(\PageIndex{2a}\) if he exerts a constant force of 75.0 N at an angle 35 below the horizontal and pushes the mower 25.0 m on level ground? The segments of the paths are the sides of a right triangle, so the path lengths are easily calculated. The work done depends on the square of the displacement, which is the same for [latex]x=\pm6\,\text{cm}[/latex], so the magnitude is 0.54 J. [/latex] For this object sliding along the surface, kinetic friction [latex]{\mathbf{\overset{\to }{f}}}_{\text{k}}[/latex] is opposite to [latex]d\mathbf{\overset{\to }{r}},[/latex] relative to the surface, so the work done by kinetic friction is negative. In many cases, it is convenient to express the dot product for gravitational work in terms of the x-, y-, and z-components of the vectors. The component of the force parallel to the displacement is the work done, as shown in the equation in the figure. Usually, there are several ways to do this, which may be more or less convenient, depending on the particular case. Forces can vary as a function of position, and displacements can be along various paths between two points. From the properties of vectors, it doesnt matter if you take the component of the force parallel to the displacement or the component of the displacement parallel to the forceyou get the same result either way. ), The part of the contact force on the object that is parallel to the surface is friction, \(\vec{f}\). (b) You dont like the new position, so you move the couch straight back to its original position (B to A in Figure \(\PageIndex{3}\)). Therefore, the work done by it is [latex]{W}_{\text{fr}}=\text{}{f}_{K}d[/latex], where d is the path length traversed. The gravitational force between two objects is an attractive force, which does positive work when the objects get closer together. 9 I have to talk about negative friction for my oral exam and I would like you to comment my presentation. In Figure \(\PageIndex{2c}\), where the person in (b) is walking horizontally with constant speed, the work done by the person on the briefcase is still zero, but now because the angle between the force exerted and the displacement is 90 (\(\vec{F}\) perpendicular to \(\vec{d}\)) and cos 90 = 0. (There are quotation marks around area because this base-height product has the units of work, rather than square meters.). Near the surface of Earth, the gravitational force on an object of mass m has a constant magnitude, mg, and constant direction, vertically down. Since the work done by a spring force is independent of the path, you only needed to calculate the difference in the quantity \(\frac{1}{2}\)kx2 at the end points. The components of a vector can be positive, negative, or zero, depending on whether the angle between the vector and the component-direction is between 0 and 90 or 90 and 180, or is equal to 90. [latex]F(x)=a[\frac{x+9\,\text{m}}{9\,\text{m}}-{(\frac{9\,\text{m}}{x+9\,\text{m}})}^{2}][/latex]. (c) What is the total work done on the lift? Note that the unstretched position is only the same as the equilibrium position if no other forces are acting (or, if they are, they cancel one another). The total displacement of the ball is zero, so no work is done. (c) Must the total work done on a particle always be positive? In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement of the object, relative to the floor, and do positive work. Provide an example. (b) How much work must be done against the elastic force of the bungee cord to stretch it 16.7 m? where [latex]|{l}_{AB}|[/latex] is the path length on the surface. Can Earths gravity ever be a constant force for all paths? }[/latex], [latex]x=\sqrt{y\text{/}(0.5\,{\text{m}}^{-1})}=\sqrt{(2\,\text{m})y}\,\text{and}\,dx=\sqrt{(2\,\text{m})}\times \frac{1}{2}dy\text{/}\sqrt{y}=dy\text{/}\sqrt{(2\,{\text{m}}^{-1})y}. Lets consider the work done by these forces in general. When you push on the wall, this feels like work; however, there is no displacement so there is no physical work. Find the work done by the same force in Example 7.4 over a cubic path, y=(0.25m2)x3y=(0.25m2)x3, between the same points A=(0,0)A=(0,0) and B=(2m,2m).B=(2m,2m). The work done by a force is the integral of the force with respect to displacement along the path of the displacement: The vectors involved in the definition of the work done by a force acting on a particle are illustrated in Figure. You can pull the rug out from under an object in such a way that it slides backward relative to the rug, but forward relative to the floor. What was the total work done against gravity, moving the book away from its original position on the shelf and back again? We have just seen that the work done by a constant force of gravity depends only on the weight of the object moved and the difference in height for the path taken, [latex]{W}_{AB}=\text{}mg({y}_{B}-{y}_{A})[/latex]. Alternately, gravity does positive work on an object that moves downward (yB

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