DOCSLIB.ORG

Work and Kinetic Energy

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Work and Kinetic Energy Work and Kinetic Energy Level : Physics I Teacher : Kim Objective Establish the relationship between work and energy Practice using Work-Kinetic Energy theorem compared to using ΣF=ma Understand how work&energy leads to the concept of simple machinery we use everyday Observe video clip on ‘Work’ Introduction of ‘Work’ done by Applied Force - Work(W) is defined as ‘force applied in the direction of motion × displacement’ W=Fcosφ × d [J] Fapp φ d= distance traveled - Both force and distance is directly related to work. - Unit of work is expressed in Joules(J) - φ is the angle the between the direction of force and the direction of motion - If you apply 1N of force over a distance of 1meter, then you are doing 1Joule of work. Q1) A boy exerts a horizontal force of 10N on a box at a distance of 25m on a flat surface. How much work did he do on the box? a) 150J b) 200J c) 250J d) 300J Q2) A girl exerts a force of 35N at an angle of 35° above the horizontal along the handle of her wagon as she pulls it 25m across a level sidewalk. How much work does she do on her wagon? a) 698J b) 717J c) 756J d) 782J Lifting a barbell is doing work. Two things happen when lifting a barbell i) a force is exerted on the barbell, and ii) the barbell is moved by that force Here, work is done on the barbell, since there is force being applied on the barbell and the barbell moves a certain distance Q3) A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J Q4) If the man is just holding the barbell above his head, how much work is done on the barbell? a) He is doing no work on the barbell. Since W=F×d and distance covered is zero, so W=0. b) Of course he is doing work, since holding the barbell above his head causes strain on his body Work done by other Forces - Any force that may exist on an object has the ‘potential’ to do work. In the figure below, three forces are acting on the block; FN, Fg and Fapp. But only Fapp contributes to doing work on the block. (surface is frictionless) FN Fapp φ d= distance traveled Fg=mg => Work done by normal force FN is zero; W=FN cos90º × d =0 => Work done by gravitational force Fg(=weight) is zero; W=Fg cos90º × d = 0 *~ The only time a force does work on an object is when the force ‘plays a role’ in moving the object Work done by Gravity i) Vertical Motion Q5) When an object is free-falling in a vacuum, what force is doing work on that object? Fg ii) Incline Plane Q6) When an object is sliding down on the FN frictionless ramp, what force is doing work on d that object? Fg sinθ Fg θ *~Work-Kinetic Energy Theorem If an object is in motion, then it must have kinetic energy(K) K= mv2 1 2 If you apply a force over a certain distance, there will be a change in kinetic energy vi distance ‘d’ vf Fapp Fapp Ki + Fcos × d = Kf 2 2 mvi +Fcosφ × d = mvf 1 1 2 2 Ki + W = Kf : Work-Kinetic Energy Theorem - Unit of energy or work is Joules[J] *~ Are we interested in the ‘acceleration’ or the ‘speed’??~* 2 W=Fcosφ × d K= mv Ki + W = Kf 1 2 A. On a Flat Surface Q7) If a 5kg object is sliding on a frictionless surface with a constant speed of 10m/s, i) What is the kinetic energy of the object? ii) If you apply a horizontal force of 5N over a distance of 10m, what is the new kinetic energy? Use ΣF=ma and Work-Kinetic Energy theorem and compare the process and result Newton’s 2nd Law : ΣF=ma Work-Kinetic Energy theorem 2 2 *~ You need to also use vf -vi =2ad Q8) A 4kg object is sliding on a frictionless surface with a constant speed of 8m/s, and you apply a horizontal force of 8N over a distance of 12m. i) How much work was done by the applied force? a) 96J b) 82J c) 74J d) 68J ii) What is the new kinetic energy? a) 162J b) 178J c) 195J d) 224J iii) What is the new speed? a) 10.6m/s b) 8.4m/s c) 7.1m/s d) 5.9m/s 2 W=Fcosφ × d K= mv Ki + W = Kf 1 2 Q9) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you apply a horizontal force of Fp=6N, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s Fp Q10) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you apply a force of Fp=6N at an angle of 27°above the surface, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s Fp 27° Q11) An object of mass 2kg is sliding on a frictionless surface with a constant speed of 8m/s. How much force Fp is needed to stop the object over a distance of 10m? (*~remember that the force is applied in the opposite direction of the motion for the object to stop~*) a) 3.9N b) 6.4N c) 8.3N d) 11.8N FN direction of motion Fp Fg 2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 Q12) A 2.5kg box is resting on the flat rough surface. The coefficient of kinetic friction between the box and the floor is µk =0.52.The box is then pulled by a force of Fp=18N. FN direction of motion fk Fp Fg i) How much work was done on the box by the pulling force Fp through a distance of 5.2m? That is, find Wp. a) 62.4J b) -62.4J c) 93.6J d) -93.6J ii) How much work was done on the box by the frictional force fk through a distance of 5.2m? That is, find Wf. a) 66.25J b) -66.25J c) 25.48J d) -25.48J iii) What is the net work W done on the box? That is, Wp + Wf. a) 27.4J b) 33.5J c) 49.7J d) 75.14J iv) What is the speed of the box when it covered 5.2m? a) 39.8m/s b) 27.2m/s c) 15.6m/s d) 4.7m/s Q13) A 4kg box is sliding on a rough surface with an initial speed of 8m/s. The coefficient of kinetic friction between the box and the floor is µk =0.4. What is the speed of the box after it moved a distanced of 4m? (*~frictional force f=µFN, where FN=mg in this case ~*) a) 19.6m/s b) 14.2m/s c) 10.5m/s d) 5.7m/s FN direction of motion fk Fg 2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 B. Vertical Motion Use Work-Kinetic Energy to solve Q3) again! A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J C. Incline Plane Q14) A 8kg box is released from rest on a 30° inclined frictionless plane and accelerates down the incline. The distance from the box to the end of the ramp is 4m i) Find the work done by gravity through a distance of 4m down the ramp. a) 271.6J b) 156.8J c) 101.2J d) 59.7J ii) What is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s Q15) A 13.2kg box is released from rest on an 11° inclined frictionless plane and accelerates down the incline. If the distance from the box to the end of the ramp is 5.1m, what is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s 2 W=Fcosφ × d f=μFN K= mv Ki + W = Kf 1 2 Q16) A 6.3kg box is being pulled with a force of 56N from the bottom of the ramp starting from rest. The surface of the ramp is frictionless and the angle of the ramp is 40°. Fp i) How much work was done on the box by the pulling force Fp through a distance of 5.2m? That is, find Wp. a) 291.2J b) 233.9J c) 187.2J d) 105.6J ii) How much work was done on the box by gravitational force Fg through a distance of 5.2m? That is, find Wg.
Load more

Recommended publications
  • Numerical Methods Applied Id Metallurgical Processing
    NUMERICAL METHODS APPLIED ID METALLURGICAL PROCESSING by JUAN HECTOR BIANCHI A thesis submitted for the degree of Doctor of Philosophy of the University of London and for the Diploma of Membership of the Imperial College John Percy Research Grou Department of Metallurgy Royal School of Mines, Imper i a 1 Co 11ege London JULY 1983 In bulk forming operations the plastic deformation is very large compared with the elastic one. This fact allows the use of constitutive laws based upon both current stress-strain rate measures. A Finite Element formulation of the mechanics involved in quasi-steady state conditions was implemented, introducing the incompressibi1ity via a penalty on the volumetric strain rate. The perfectly plastic behaviour was considered first, and the extrusion process was thoroughly analyzed. Solutions for direct and and indirect modes of operation up to very high extrusion ratios are presented. Both plane strain and axisymmetric geometries were considered. Frictional conditions at the billet-container interface were introduced in two ways. Comparison between results obtained using these lines of approach and numerical problems associated to them are examined. Pressure results are compared with previously reported solutions resulting from Slip Lines, Upper Bounds and Finite Elements. The behaviour of the internal mechanics as a function of changes of geometrical and frictional conditions are examined. Next, a non-linear flow stress resulting from hot torsion experi- ments was considered. The thermal analysis involves the solution of another equation, which is coupled with the mechanical problem due to convection, heat generation and flow stress dependence on temperature. After an appropriate transformation, that equation is formulated in a "weak" form and implemented to be solved jointly with the mechanical part.
    [Show full text]
  • Experiment 4: Newton's 2Nd
    Experiment 4: Newton's 2nd Law - Incline Plane and Pulley In this lab we will further investigate Newton's 2nd law of motion by using an incline-pulley system. The incline-pulley system, shown in Figure 1, can be classified as a simple machine, that is, one of the classic elementary devices that more complicated and advanced machines are built around. As shown in Figure 1, the acceleration of the mass along the inclined plane (M1) can be controlled by using a hanging counterweight (M2) over the pulley and/or varying the angle of the incline. The free body diagrams for the two masses are shown in Figure 2. We will use the airtrack to create a frictionless plane and also assume that the pulley is frictionless with uniform tension in the string. With these assumptions, the acceleration P of the two masses are the same (a1;x = a2;y). Applying Newton's second law, F = ma, to the freebody diagram, we can write a system of equations describing the motion of the two masses (T is the tension in the string): m1a = T − m1gsinθ (1) m2a = m2g − T (2) Solving these equations for the acceleration: (M − M sin(θ))g a = 2 1 (3) M1 + M2 M1 M2 θ Figure 1: A mass M1 slides along a frictionless incline of angle θ with a counterweight M2 passing over a pulley. 1 a1,x a2,y T N T y x y θ x W1 = m1g W2 = m2g Figure 2: Freebody diagram for the two masses. Experimental Objectives The objective of the lab is to experimentally test the theoretical acceleration (Eq.
    [Show full text]
  • Rolling Motion: Experiments and Simulations Focusing on Sliding Friction Forces
    Rolling motion: experiments and simulations focusing on sliding friction forces Pasquale Onorato, Massimiliano Malgieri and Anna De Ambrosis Dipartimento di Fisica Università di Pavia via Bassi, 6 27100 Pavia, Italy Abstract The paper presents an activity sequence aimed at elucidating the role of sliding friction forces in determining/shaping the rolling motion. The sequence is based on experiments and computer simulations and it is devoted both to high school and undergraduate students. Measurements are carried out by using the open source Tracker Video Analysis software, while interactive simulations are realized by means of Algodoo, a freeware 2D-simulation software. Data collected from questionnaires before and after the activities, and from final reports, show the effectiveness of combining simulations and Video Based Analysis experiments in improving students’ understanding of rolling motion. Keywords Rolling motion, friction force, collisions, video analysis, simulations, and billiard. Introduction As it is well known, rolling motion is a complex phenomenon whose full comprehension involves the combination of several fundamental physics topics, such as rigid body dynamics, friction forces, and conservation of energy. For example, to deal with collisions between two rolling spheres in an appropriate way requires that the role of friction in converting linear to rotational motion and vice-versa be taken into account (Domenech & Casasús, 1991; Mathavan et al. 2009; Wallace &Schroeder, 1988). In this paper we present a sequence of activities aimed at spotlighting the role of friction in rolling motion in different situations. The sequence design results from a careful analysis of textbooks and of research findings on students’ difficulties, as reported in the literature.
    [Show full text]
  • Physics 20 Homework 3 SIMS 2016
    Physics 20 Homework 3 SIMS 2016 Due: Thursday, August 25th Special thanks to Sebastian Fischetti for problems 1, 5, and 6. Edits in red made by Keith Fratus. 1. The ballistic pendulum is a device used to measure the speed of a bullet. A bullet of mass m is fired into a block of mass M hanging on a pendulum of length L, into which it embeds itself; this causes the pendulum to swing up to some maximum angle θ, as shown. Calculate the initial speed of the bullet in terms of m, M, θ, L, and g. Hint: What conservation laws might you be able to apply in this problem? When can you apply each one, and when can you not? Figure 1: The operation of a ballistic pendulum. Special thanks to Sebastian Fischetti for the figure. 1 2. In lecture, we discussed the definition of potential energy for a single particle which is subjected to an external force. However, we know that in reality, forces arise from the interactions between multiple particles. It is possible to generalize the definition of potential energy to describe two particles interacting with each other as a closed system. As an example, the potential energy of a pair of neutral atoms can be modelled, to a very good approximation, by the Lennard-Jones potential, given by r 12 r 6 U(r) = U 0 − 2 0 0 r r Here U0 and r0 are constants, and r is the distance between the two atoms. Despite the fact that the atoms are electrically neutral overall, the charge due to the protons and electrons is spread out over a non-zero region, so there is still a small, residual electrical interaction between the two.
    [Show full text]
  • 5 | Newton's Laws of Motion 207 5 | NEWTON's LAWS of MOTION
    Chapter 5 | Newton's Laws of Motion 207 5 | NEWTON'S LAWS OF MOTION Figure 5.1 The Golden Gate Bridge, one of the greatest works of modern engineering, was the longest suspension bridge in the world in the year it opened, 1937. It is still among the 10 longest suspension bridges as of this writing. In designing and building a bridge, what physics must we consider? What forces act on the bridge? What forces keep the bridge from falling? How do the towers, cables, and ground interact to maintain stability? Chapter Outline 5.1 Forces 5.2 Newton's First Law 5.3 Newton's Second Law 5.4 Mass and Weight 5.5 Newton’s Third Law 5.6 Common Forces 5.7 Drawing Free-Body Diagrams Introduction When you drive across a bridge, you expect it to remain stable. You also expect to speed up or slow your car in response to traffic changes. In both cases, you deal with forces. The forces on the bridge are in equilibrium, so it stays in place. In contrast, the force produced by your car engine causes a change in motion. Isaac Newton discovered the laws of motion that describe these situations. Forces affect every moment of your life. Your body is held to Earth by force and held together by the forces of charged particles. When you open a door, walk down a street, lift your fork, or touch a baby’s face, you are applying forces. Zooming in deeper, your body’s atoms are held together by electrical forces, and the core of the atom, called the nucleus, is held together by the strongest force we know—strong nuclear force.
    [Show full text]
  • Dynamics of Rigid Bodies
    EN 4 Dynamics and Vibrations Homework 10: Rigid Body Dynamics Due: WEDNESDAY, April 29, 2009 5:00pm 1. Galileo’s Classic Problem A cylindrical object having mass m, radius R, and moment of inertia I rolls down an incline of angle θ without slipping. What is the acceleration of the center of mass down the hill, as a function of m, I, g, and θ? Evaluate the result in the following cases: (a) the mass is concentrated at the center of mass connected to a massless cylindrical frame (b) all the mass is concentrated at the outer rim of the cylinder (c) the mass is equally distributed throughout the cylinder. (d) How do these cases compare to the acceleration of a point particle of mass m sliding down a frictionless plane? (e) For case (c), if the object starts from rest, what is the velocity of the center of mass after rolling a distance d down the incline? (f) For case (c), now use the work-energy for relation for rigid bodies to determine the velocity after rolling a distance d down the incline. Remember that “rolling without slipping” means that the friction at the contact point is static friction (the contact point has zero velocity) and so there is no energy lost due to this static friction (energy loss only occurs when there is slipping). θ 2. Yo-Yo Motion A Yo‐Yo of mass m and moment of inertia 2 IG=mk , where k is a distance called the “radius of gyration”, moves by “rolling” along a string wound on the inner radius r, as shown.
    [Show full text]
  • Kinetic Energy 327 7 | WORK and KINETIC ENERGY
    Chapter 7 | Work and Kinetic Energy 327 7 | WORK AND KINETIC ENERGY Figure 7.1 A sprinter exerts her maximum power to do as much work on herself as possible in the short time that her foot is in contact with the ground. This adds to her kinetic energy, preventing her from slowing down during the race. Pushing back hard on the track generates a reaction force that propels the sprinter forward to win at the finish. (credit: modification of work by Marie-Lan Nguyen) Chapter Outline 7.1 Work 7.2 Kinetic Energy 7.3 Work-Energy Theorem 7.4 Power Introduction In this chapter, we discuss some basic physical concepts involved in every physical motion in the universe, going beyond the concepts of force and change in motion, which we discussed in Motion in Two and Three Dimensions and Newton’s Laws of Motion. These concepts are work, kinetic energy, and power. We explain how these quantities are related to one another, which will lead us to a fundamental relationship called the work-energy theorem. In the next chapter, we generalize this idea to the broader principle of conservation of energy. The application of Newton’s laws usually requires solving differential equations that relate the forces acting on an object to the accelerations they produce. Often, an analytic solution is intractable or impossible, requiring lengthy numerical solutions or simulations to get approximate results. In such situations, more general relations, like the work-energy theorem (or the conservation of energy), can still provide useful answers to many questions and require a more modest amount of mathematical calculation.
    [Show full text]
  • Sample Lab Report
    Name: Experiment Example, Acceleration due to Gravity The purpose of this experiment is to measure the acceleration due to gravity. We will use a nearly-frictionless inclined plane. The acceleration of a point particle sliding freely down an inclined plane is given by a=qgsin By varying the angle of the incline and measuring the acceleration, the acceleration due to gravity g can be determined. We will use a smart-pulley attached to a CBL to determine the acceleration of the object. By graphing the acceleration versus sinq, the acceleration due to gravity can be determined. The general set-up is shown below. Smart Pulley attached to CBL In your write-up, please include the following sections: (1) Title, Names, etc. (Section 1 on the handout: Lab Reports) (2) Purpose (Section 2 on the handout: Lab Reports) (3) Method (Section 3 on the handout: Lab Reports) (4) Data (Section 4 on the handout: Lab Reports) You will want to think about how much data you will require. Obviously, more data is better; however, you have a limited amount of time. Make sure you obtain enough data points and trials to create a good graph. (5) Analysis (Section 5 on the handout: Lab Reports) Include a graph of Acceleration versus sinq. Include an appropriate regression equation with R2 value. Determine the acceleration due to gravity. (6) Conclusion (Section 6 in the handout: Lab Reports) Your conclusion will need to be a couple of good paragraphs. You will want to consider the following questions: What factors affected your result for the acceleration due to gravity? How does your value for g compare with the accepted value? Quantitatively compare your value of g with the accepted value.
    [Show full text]
  • Force and Laws of Motion 115
    Chapter 9 FFFORCEORCEORCE ANDANDAND LAWSAWSAWS OFOFOF MOTIONOTIONOTION In the previous chapter, we described the In our everyday life we observe that some motion of an object along a straight line in effort is required to put a stationary object terms of its position, velocity and acceleration. into motion or to stop a moving object. We We saw that such a motion can be uniform ordinarily experience this as a muscular effort or non-uniform. We have not yet discovered and say that we must push or hit or pull on what causes the motion. Why does the speed an object to change its state of motion. The of an object change with time? Do all motions concept of force is based on this push, hit or require a cause? If so, what is the nature of pull. Let us now ponder about a ‘force’. What this cause? In this chapter we shall make an is it? In fact, no one has seen, tasted or felt a attempt to quench all such curiosities. force. However, we always see or feel the effect For many centuries, the problem of of a force. It can only be explained by motion and its causes had puzzled scientists describing what happens when a force is and philosophers. A ball on the ground, when applied to an object. Pushing, hitting and given a small hit, does not move forever. Such pulling of objects are all ways of bringing observations suggest that rest is the “natural objects in motion (Fig. 9.1). They move state” of an object.
    [Show full text]
  • Type III Inclined Planes, Hills, Ramps
    Type III ­ Inclined Planes, Hills, Ramps (printed copy for students) + y *Recall that if there is a force of friction it acts in the opposite direction of the motion. FN *The coordinate system is rotated the same amount as the angle θ. y + x Fgx plane of ramp ­ x θ Fgy θ "g" stands for gravity. Fg ­ y Fgy and Fg are separated by θ because of two similar triangles. Fgx = Fgsin θ component parallel to the plane. Fgy = Fgcos θ component perpendicular to the plane. NOTE! The sin and cos have switched places. This will only happen when dealing with objects on a ramp. NOTE FURTHER! Every F in the above diagram can be replaced with an a for acceleration. 1. A 2300 kg block is sliding down a hill inclined at an angle of 35o. The coefficient of kinetic friction is 0.13. What is the acceleration of the car? + y ­ x + x ax = ? ­ y 35o 2 ay ag = 9.81 m/s Examples 2. A skier coasts down a 3.5o slope at a constant speed. Find the coefficient of kinetic friction between the skis and the snow covering the slope. (0.061) + y + x *positive direction is up the ramp. ­ x ­ y force = mass x acceleration af = acceleration from frictional force. aN = acceleration from normal force. "y" direction negative vertical direction constant speed "x" direction negative horizontal direction acts up the ramp, so is positive Physics 12 Sample Problems ­ Inclined Planes Handout 1. A trunk weighing 562 N is resting on a plane inclined at 30.0o from the horizontal.
    [Show full text]
  • The Lagrangian Method Copyright 2007 by David Morin, [email protected] (Draft Version)
    Chapter 6 The Lagrangian Method Copyright 2007 by David Morin, [email protected] (draft version) In this chapter, we're going to learn about a whole new way of looking at things. Consider the system of a mass on the end of a spring. We can analyze this, of course, by using F = ma to write down mxÄ = ¡kx. The solutions to this equation are sinusoidal functions, as we well know. We can, however, ¯gure things out by using another method which doesn't explicitly use F = ma. In many (in fact, probably most) physical situations, this new method is far superior to using F = ma. You will soon discover this for yourself when you tackle the problems and exercises for this chapter. We will present our new method by ¯rst stating its rules (without any justi¯cation) and showing that they somehow end up magically giving the correct answer. We will then give the method proper justi¯cation. 6.1 The Euler-Lagrange equations Here is the procedure. Consider the following seemingly silly combination of the kinetic and potential energies (T and V , respectively), L ´ T ¡ V: (6.1) This is called the Lagrangian. Yes, there is a minus sign in the de¯nition (a plus sign would simply give the total energy). In the problem of a mass on the end of a spring, T = mx_ 2=2 and V = kx2=2, so we have 1 1 L = mx_ 2 ¡ kx2: (6.2) 2 2 Now write µ ¶ d @L @L = : (6.3) dt @x_ @x Don't worry, we'll show you in Section 6.2 where this comes from.
    [Show full text]
  • Chapter 5 NEWTON's LAWS of MOTION
    Nerd Bling UNIVERSITY PHYSICS Chapter 5 NEWTON’S LAWS OF MOTION Monday: • Newton’s Three Laws of Motion • FREE BODY DIAGRAMS • Inclined planes Wednesday: • Action-Reaction forces • Pulleys and Ropes Man of the Millennium Sir Issac Newton 1687 Published Principia •Invented Calculus (1642 -1727) •3 Laws of Motion •Universal Law of Gravity What Is a Force? . A force is a push or a pull. A force acts on an object. Pushes and pulls are applied to something. From the object’s perspective, it has a force exerted on it. Contact forces are forces that act on an object by touching it at a point of contact. Long-range forces are forces that act on an object without physical contact. (gravity, electricity,etc) Slide 5-18 Forces are Interactions FFEarth on Rock Rock on Earth Thinking About Force . Every force has an agent which causes the force. Forces exist at the point of contact between the agent and the object (except for the few special cases of long-range forces). Forces exist due to interactions happening now, not due to what happened in the past. Consider a flying arrow. A pushing force was required to accelerate the arrow as it was shot. However, no force is needed to keep the arrow moving forward as it flies. It continues to move because of inertia. Slide 5-75 Galileo Challenged The Dogma Of Aristotle’s Natural Motion (Forces are needed for any motion) The natural motion of a body is to remain in whatever state of motion it is in unless acted upon by net external forces.
    [Show full text]