TEXTBOOK REFERENCES: College Physics a strategic approach (AP Edition) (Fourth Edition) Chapter 10

AP Physics-B Work / Power / Eneregy INTRODUCTION: In everyday conversation, the words work and power are used to describe many activities. In physics, the term work has a precise definition it is not merely an activity where one exerts an effort. Work is a physical quantity. The amount of work done can be calculated. Power is also a physical quantity. It is not merely a measure of one s strength, but the rate at which work is done. The purpose of this unit is to familiarize you with the physics terms work and power. WORK is defined as the product of the applied force and the displacement through which an object moves in the direction of the applied force. Work is done only when an object moves some distance due to the net force. An applied force causing no acceleration or motion at a constant velocity means no net work is done on the body. In vector terms, WORK is defined as the dot product of the applied force and the displacement. WORK is a scalar quantity even though force and displacement are vector quantities. The units for work are NEWTON-METER or JOULE. W = F ΔX PERFORMANCE OBJECTIVES: Upon completion of the activities and readings in this unit and when asked to respond either orally or on a written test, you will: state the physical definition for work. Solve problems using the equation for work. know that the SI unit for work is a JOULE. recognize that work is a scalar quantity. recognize that the equation for work is not as simple as it first seems. Be able to solve problems where the force is at an angle with the displacement. Be able to solve problems where the applied force is a function of the displacement. state the definition for power. Write the equation for power. correctly calculate power is watts and in kilowatts demonstrate an understanding of the relationship between work and energy transfer. Recognize the usefulness of the term work. define a simple machine. Know that simple machines are of two basic types. Be able to calculate the mechanical advantage and efficiency of a machine. Define work as the product of force and displacement and in terms of energy transfer. Define kinetic energy. Calculate the kinetic energy of a body given the mass and velocity of the body. Calculate the change in kinetic energy of a body given the force applied and the displacement. Define potential energy. Calculate the gravitational potential energy of a body near the surface of the earth. State the law of conservation of energy. Define elastic collision. Use the conservation of energy and momentum to solve collision problems in one dimension. Solve problems with equal mass colliding elastically at an angle. Define inelastic collision. Describe energy changes in a pendulum swing. Solve problems associated with a pendulum. TEXTBOOK REFERENCES: College Physics a strategic approach (AP Edition) (Fourth Edition) Chapter 10 Work is man s great function. He is nothing, he can do nothing, he can achieve nothing, fulfill nothing without working. If you are poor work. If you are rich, continue working. If you are burdened with seemingly unfair responsibilities, work! If you are happy, keep right on working! Idleness gives room for doubts and fears. If disappointment comes, work! If your health is threatened work! When faith falters work! When dreams are shattered and hope seems dead work! Work as if your life were in peril, it really is. No matter what ails you work! Work faithfully work with faith. Work is the greatest remedy available for both mental and physical afflictions! - James M. Cohan QUESTIONS AND PROBLEMS: 1.) A force of 100.0 N is needed to push a car across an empty parking lot at a constant speed. Two students push the car 40.0 m. How much work is done by the students? There is obviously another force acting on the car, since the net force is zero. How much work is done by this other force? c.) What is the total net work done on the car? 4000 J -4000 J zero 2.) a.) What is the weight of a 49 kg crate? b.) What work is done to lift the crate a distance of 10.0 m? 480.2 N 4802 J 3.) A package weighs 35 N. A person carries the package from the ground floor to the fifth floor of an office building, or 15 m upward. a.) How much work does the person do on the package? If the person weighs 750 N, How much total work is done on the package and the person? 525 J 11,775 J

4.) A worker carries cement blocks, weighing 150 N each up a ladder onto a scaffold 8.0 m high. The worker carries them at a rate of 2 blocks per minute. How much work is done by the worker in a.) 10 minutes? b.) one hour? 24,000 J 1.4 x 10 5 J 5.) A person carries a bag of groceries weighing 100 N a horizontal distance of 50 m. How much work is done on the groceries? 6.) In driving to the top of a mountain, the amount of work done is about the same whether the road goes straight up of zigzags in gradual slopes. Why then do most mountain roads wind up gradually rather than going straight up? 7.) The design of the road dividers that separate opposing traffic lanes used to consist of metal rails mounted to make contact with the fenders and the doors of the car. The newer design is of concrete with a cross-section very deliberately shaped to make contact with the tires. Why is this new design more effective in slowing down a car that side-swipes it? 8.) Scissors for cutting paper have long blades and short handles, whereas metal-cutting shears have long handles and short blades. Bolt cutters have very long handles and very short blades. Why is this so? SIMPLE MACHINES: Machines enable a person to do some work by changing the amount, the direction, or the point of application of the necessary force. Many times, a machine is used to do work on a body using less force but applying the force through a greater displacement. The work output of a machine can never be greater than the work input. Ideally input and output would be the same, but in real life situations, the output is less than the input. The ratio of work output to work input is called the efficiency of the machine. The ratio of the output force exerted by the machine on a load to the input force exerted by the operator on the machine is defined as the actual mechanical advantage (AMA). The ideal mechanical advantage (IMA) is defined as the ratio of the distance through which the input force acts to the distance through which the output force acts. Efficiency is also equal to the ratio of the AMA to the IMA. Three basic types of simple machines are the lever, the inclined plane, and the pulley. Eff = Work output / Work input = AMA / IMA IMA = ΔX input / ΔX output AMA = F output / F input 9.) Workers use a force of 2125 N to push a piano weighing 8800 N up a 20.0 m long ramp. a.) How much work is done on the piano by the workers in pushing it up the ramp? b.) The piano is being moved from street level to the second floor of a building. The second floor is 4.0 m above street level. If the workers decide to lift the piano straight up by using ropes, how much work would they do? c.) What is the efficiency of the ramp? 42,500J 35,200 J 83 % 10.) In order to change a tire, a force of 80.0 N is exerted on the handle of a screw-type bumper jack. The handle to which the steel shaft is attached has a radius of 0.50 m. The handle is turned through 30 revolutions. How much work is done? 7540 J 11.) Using a pulley system, a worker does 1920 J of work to lift a crate from one floor to the next in a warehouse. The force is exerted through a distance of 40.0 m on the rope of the pulley system, which is 100% efficient. How much force was used to lift the crate? 48 N 12.) A force of 3.0 N is required to raise a load of 16 N by means of a pulley system. If the load is raised 1.0 meters while the applied force is exerted through a distance of 8.0 m, find a.) the IMA b.) the AMA, and c.) the efficiency of the pulley system. 8 5.3 67% WORK AND THE DIRECTION OF THE FORCE: In calculating work, use only the component of the force that acts in the direction of the motion. For example, if you push a lawn mower, the vertical component of the force is balanced by the upward push of the ground. The force that is doing the work is the horizontal component, Fx. The work done by a force acting at some angle with the motion is the product of the force, the displacement, and the cosine of the angle between them. Work = F(cosΘ)ΔX 13.) A person applied a force of 600 N to a rope which is tied to a metal box. The rope is held at an angle of 45 with the floor while pulling the box across the floor. The box is moved a distance of 15 m at a constant speed. How much work is done on the box? 6364 J

14.) A loaded sled weighing 800.0 N is pulled a distance of 200.0 m. To do this, a force of 120 N is exerted on a rope that makes an angle of 60 with the horizontal. How much work is done on the sled? 12000 J 15.) Because of friction, a force of 400 N is needed to drag a wooden crate across a floor. The rope tied to the crate is held at an angle of 53 with the horizontal. a.) How much tension is needed in the rope to move the crate? b.) What work is done if the crate is dragged 25 m? 665 N 10,000 J 16.) A cable attached to a small tractor pulls a barge through a canal lock. The tension in the cable is 2500 N. It makes an angle of 30 with the direction in which the barge is moving. a.) What force moves the barge along the lock? b.) If the lock is 200.0 m long, how much work is done to get the barge through the lock? 2165 N 433,000 J 17.) It takes 17,600 J of work to pull a loaded sled weighing 2200 N a distance of 22 m. To do this, a force of 1600 N is exerted on a rope that makes an angle with the horizontal. At what angle is the rope held? 60 WORK AND VARYING FORCE: Up to this point, the forces applied were constant, that is the same amount of force was applied for the entire displacement. Sometimes, the force applied is a function of the displacement. A good example of this is when you compress or stretch a spring the force increases with displacement. In calculus terms, the definition for work is the integral of the dot product. This means to find the area under the force-displacement curve. In the next few problems, we will look at work done on springs. 18.) Sketched at the right is a force vs. compression graph for a spring. How much work is done to compress the spring 0.3 meters? about 0.45 J 19.) The force-compression curve of a spring is shown at the right. How much work is done in compressing the spring 0.3 meters? 5.0 J 20.) A linear elastic spring is compressed by 0.2 meters by a force of 20 Newtons. How much work is done to compress the spring? about 2.0 J POWER: is the rate of doing work: work per unit time. The SI unit of power is the watt. A watt is one joule per second. A kilowatt is 1000 watts. One horsepower = 746 watts. Power (P) = Work (W) / Δt 21.) A machine produces a force of 40.0 N through a distance of 100.0 meters in 5.00 seconds. a.) How much work is done? b.) What is the power of the machine in watts? c.) in kilowatts? 4000J 800 W 0.8 kw 22.) A box that weighs 1000.0 N is lifted a distance of 20.0 m straight up by a rope and pulley system. The work is done in 10 seconds. What amount of power is used in a.) watts? b.) kilowatts? 2000 W 2 kw 23.) A hiker carries a 20.0 kg knapsack up a trail. After 30.0 minutes, he is 300.0 m higher than his starting point. a.) What is the weight of the knapsack? b.) How much work is done? c.) If the hiker weighs 600.0 N, how much total work is done? d.) During the 30 minutes, what is the hiker s average power in watts? e.) kilowatts? 196 N 58,800 J 238,800 J 132 W 0.132 kw 24.) How much work does a 400.0 watt motor do in 5.0 minutes? 120,000 J

25.) A pump raises 30 liters of water per minute from a depth of 100.0 m. What is the wattage expended? (a liter of water has a mass of 1.0 kg) 490 W When a constant force acts on a body in the direction of the body s motion, the power can be expressed as the dot product of force times velocity. Power = F v 26.) A motor operates an endless conveyor belt. The motor produces a force of 500.0 N on the belt and it moves at the rate of 6.5 m/s. What is the power of the motor in kilowatts? 3.25 kw REVIEW: 27.) Water falls at a rate of 8000 kg/s from the top of a large dam and strikes the blades of a turbine 50.0 meters below. If 80% of the falling water effectively does work on the turbine, what electric power in megawatts can the turbine develop? 3.1 MW 28.) Using a 98 N force, a 20.0 kg mass is slid up a frictionless incline a distance of 10.0 m. a.) How much work is done on the mass? b.) How much work is done if the 20.0 kg mass is lifted straight up 5.0 m? 980 J 29.) A 70.0 N force moves a 10.0 kg mass 25 meters up a 37 incline. a.) How much work is done on the object? b.) How much work is done if the object is lifted straight up? c.) Explain the apparent discrepancy? 1750 J 1470 J 30.) A loaded elevator weighs 1.2 x 10 4 N. An electric motor hoists the elevator 9.0 m in 15 s. a.) How much work is done? b.) What is the power in watts and kilowatts? 1.08 x 10 5 J 7.2 x 10 3 W 7.2 kw 31.) A truck s engine delivers 90,000 watts of power. The truck must overcome 4500 N of force that resists the motion of the truck. What is the truck s speed? 20 m/s 32.) A person applies a force of 60.0 N for a distance of 20.0 m to push a desk across a floor at a constant speed. How much work is done by the person? b.) There is obviously some other force acting on the desk since the net force is zero and the desk therefore maintains constant speed. How much work is done by this other force? c.) What is the total work done on the desk? 1200 J -1200 J zero 33.) A force of 60.0 N is needed to push a crate weighing 300 N at a constant speed across a waxed floor. How much work must be done to push the crate 15 meters? b.) The net force is zero since the crate moves with constant speed. There must be a frictional force acting between the crate and the waxed floor. How much work is done by this frictional force? c.) What is the total work done on the crate? 900 J -900 J zero 34.) What is the weight of a 50.0 kg crate? b.) To lift the crate at constant speed, a person must apply a force equal to the weight of the crate. How much work is done by the person to lift the crate 10.0 m at constant speed? c.) Since there is no acceleration, the net force on the crate is zero. What other force is acting on the crate? d.) How much work is done on the crate by the earth s gravitational force of attraction? e.) What is the total work done on the crate? 490 N 4900 J F g -4900 J zero 35.) What do you expect the crate in Problem 33 to do after the person stops pushing? b.) What do you expect the crate in Problem 34 to do if the person no longer pushes it upward? c.) Explain why your answers are the same or different. d.) Friction is called a non-conservative force and gravity is called a conservative force. Do these definitions support or contradict your answer to part c? Energy and its Conservation Introduction: Energy is a term that most of us take for granted and use quite freely. We assume we know what we are talking about when speaking of energy. In truth, energy is probably the most complex and least understood of all physical quantities. Thus, in this unit, energy will be defined as the capacity to do work. However, this definition states what energy does, not what it is. Still, as you learn the basic concepts related to energy, we will refer to it in this way. All forms of energy are interchangeable. When energy changes form the total energy of a system remains the same. Kinetic Energy: The energy an object possesses because of its motion. Kinetic energy depends on mass and velocity. K = ½mv 2. The net work done on a body can change its kinetic energy: Work net = ΔK = K f - K i 36.) An 8.0 kg mass moves at 30.0 m/s. What is its kinetic energy? 3600J

37.) An electron with a mass of 9.1 x 10-31 kg moves through a vacuum with a speed of 2.5 x 10 8 m/s. Find the electron's kinetic energy. 2.8 x 10-14 J 38.) An alpha particle travels at 1.6 x 10 7 m/s. The kinetic energy of the particle is 6.0 x 10-13 J. What is the mass of the alpha particle? 4.7 x 10-27 kg 39.) A 10.0 kg body has a kinetic energy of 500.0 J. What is the body's velocity? What is its momentum? 10 m/s 100 kg m/s 40.) A 10.0 kg body at rest on a frictionless table receives an impulse of 30.0 N s. What is its final kinetic energy? 45 J 41.) A constant 20.0 N force acts for 10.0 s on a 5.0 kg object that is initially at rest. a.) What is the object's final momentum? b.) What is the object's final kinetic energy? c.) How much work must be done in accelerating the body? 200 kg m/s 4000 J 4000 J 42.) A 10.0 kg mass is moving with constant speed of 10.0 m/s. a.) How much work must be done to double the speed to 20 m/s? b.) How much work must be done to reduce its speed to 5.0 m/s? 1500 J 375 J 43.) A car of mass 1100 kg moves at 20 m/s. A breaking force of 1200 N is needed to bring the car to a halt. a.) What is its stopping distance? b.) If the same car is moving at 40 m/s, what is its stopping distance? 183 m 732 m 44.) What is the kinetic energy of a 1600 kg car which moves at a.) 30.0 km/hr? b.) 60.0 km/hr? c.) What is the ratio of (b) to (a)? 55000 J 220000 J 4:1 45.) How does the kinetic energy of a car change when its speed is doubles? 46.) Compare the kinetic energies of two objects, A and B, identical in every aspect except one. Assume that the singe difference is: a.) Object A has twice the velocity of B. b.) Object A moves North, B South c.) Object A moves in a circle, B in a straight line. d.) Object A is a projectile falling vertically downward, B is a projectile moving vertically upward at the same speed. 47.) Two bodies of unequal mass each have the same kinetic energy and are moving in the same direction. If the same retarding force is applied to each, how will the stopping distances of the bodies compare? 48.) A force of 30.0 N accelerates a 2.0 kg object from rest for a distance of 3.0 m along a level, frictionless surface; the force then changes to 15 N and acts for an additional 2.0 m a.) What is the final kinetic energy of the object? b.) How fast is it moving? 120 J 11 m/s Potential Energy: The energy an object has because of its position, configuration or state. The potential energy an object has due to its position above the surface of the earth is called its Gravitational Potential Energy - U G. The increase in the potential energy of any system is equal to the work done on the system. When the distance above the earth is very small compared to the radius of the earth, we consider the gravitational field to be uniform and treat the gravitational force on a body as a constant. If the lowest position is chosen as the zero position, then the work necessary to lift an object a vertical displacement above the zero position is stored as gravitational potential energy: UG = mgh: where m is the mass in kilograms, g is the acceleration due to gravity in m/s 2 and h is the height above the zero position in meters. Once again, the increase in the potential energy of any system is equal to the work done on the system. For example, if you lift a 5.0 kg box a distance of 2.0 meters above the ground, the work done on the box is 98 J. You are doing work on the box-earth system. The increase in the potential energy of the box-earth system is 98 J. You are actually working against gravity. The work done by gravity on the box is -98 J. The work done against gravity is stored as potential energy. 49.) A 5.0 kg bowling ball is lifted from the floor to a height of 1.5 m. What is the increase in its potential energy? 73.3 J 50.) A person weighing 630 N climbs up a ladder to a height of 5.0 m. a.) What work does the person do? b.) What is the increase in the potential energy of the earth-person system when the person is at this height? 3150 J

The Law of Conservation of Energy: The total energy of a system cannot change unless work is done on the system. Within an isolated system, energy always remains the same. Energy can never be "lost" by a system it can; however, change forms. Energy can be transferred between systems when one system does work on another. Work done by a conservative force conserves mechanical energy. If only conservative forces are working in a system, mechanical energy is conserved: E = K + U! This means that at any instant, the total mechanical energy (potential and kinetic) is a constant. We will study two conservative forces in this unit gravitational and elastic forces. The work done a conservative force always has these properties: 1.) It is equal to the difference between the initial and final values of a potential energy function. 2.) It is reversible. 3.) It is independent of the path of the body and depends only on the starting and ending points. 4.) When the starting and ending points are the same, that is when the body travels a closed path, the total work done by the conservative force is zero. The conservation of energy can be used to solve problems that would be more complicated if solved directly using Newton's Laws. According to the conservation of mechanical energy: At any instant, the total mechanical energy (potential and kinetic) is a constant. The total mechanical energy at the beginning of a process is equal to the total mechanical energy at the end of the process. Net Work (W net) = ΔK + -ΔU U i + K i = U f + K f You would be considered an outside force acting on a box-earth closed-system. Work done by an outside force = ΔU 51.) An 8.0 kg flowerpot falls from a window ledge 12.0 m above the sidewalk. a.) What is the potential energy of the flowerpot just before it falls? b.) What is the kinetic energy of the flowerpot just before it strikes the sidewalk? c.) Determine the speed of the pot just before it strikes the sidewalk. 940 J 940 J 15 m/s 52.) A 15.0 kg model glider flies horizontally at 12.5 m/s. a.) Calculate its kinetic energy. b.) The plane goes into a power dive and levels off 20.4 m closer to the earth. How much potential energy did it lose during the dive? c.) How much kinetic energy did the plane gain during the dive? d.) What is its new kinetic energy? e.) Neglecting frictional effects, what is its new horizontal velocity? 1171 J 3000 J 3000 J 4171 J 23.6 m/s 53.) A 10.0 kg test rocket is fired from Kennedy Space Center. Its fuel gives it a kinetic energy of 1960 J before it leaves the ramp. How high with the rocket rise? 20 m 54.) In an electronics factory, small cabinets slide down a 30 incline for a distance of 16 m to reach the next assembly stage. The cabinets have a mass of 10.0 kg each. a.) What kinetic energy would a cabinet attain when it gets to the bottom of the incline? b.) Calculate the speed each cabinet would acquire if the incline were frictionless. 784 J 12.5 m/s 55.) A ping-pong ball of mass m rolls off the edge of a table that is 1.0 m high. When the ball strikes the floor its velocity is 5.0 m/s. How fast was it rolling when it left the table? 2.3 m/s Work and the Change in Kinetic Energy When Multiple Forces are Involved: 56.) The students from problem 1 decide to see just how much force they can apply to the car that they ve been pushing on. As it turns out, they instantaneously increase their force to 136 N. How much Kinetic Energy does the car now gain as it moves the next 5 m? 180 J Elastic Collisions: An elastic collision (specifically, a perfectly elastic collision) is a collision in which the total kinetic energy of the objects involved is exactly the same before and after the collision. Strictly speaking, all collisions between objects larger than individual particles of matter are inelastic. In an inelastic collision, the total kinetic energy decreases during the collision. In some collisions between large objects, very little kinetic energy is lost and we treat these collisions as perfectly elastic collisions. The collision between two billiard balls or two glass marbles moving across a smooth surface is nearly perfectly elastic and shall henceforth be referred to as elastic. The law of conservation of momentum holds for all collisions. Although kinetic energy decreases during an inelastic collision, there is no change in momentum. In an elastic collision, the behavior of objects is predictable. This is done by considering the laws of conservation of energy and conservation of momentum together.

57.) Two bowling balls, each with a mass of 8.0 kg, roll together along a smooth ramp at 10.0 m/s. they collide with a row of 8.0 kg bowling balls at rest. Collisions between bowling balls are nearly elastic. a.) After the collision, can one bowling ball leave the opposite end of the row with a speed of 20.0 m/s and satisfy the law of conservation of momentum? b.) If one bowling ball did leave the opposite end of the row at 20.0 m/s, what kinetic energy would it possess? What was the total kinetic energy of the two balls before the collision? Would energy be conserved under these circumstances? c.) Two bowling balls leave the opposite ends of the row at 10.0 m/s. Is the law of conservation of momentum obeyed? Is the law of conservation of energy obeyed? Yes, 1600 J, 800 J, No, Yes, Yes 58.) A 10.0 kg mass moving with a velocity of 5.0 m/s right collides head-on with a stationary 10.0 kg mass. The collision is elastic. What is the final velocity of each of the masses? zero 5 m/s right 59.) A 2.0 kg mass moving at a velocity of 2.0 m/s left collides elastically head-on, with a 5.0 kg mass initially at rest. Find the final velocities of both masses. 0.857 m/s 1.143 m/s 60.) A 10.0 kg mass moving with a velocity of 5.0 m/s right collides elastically, with a 5.0 kg mass which is initially at rest. What is the final kinetic energy of the system? 125 J 61.) A 20 kg mass moving at a speed of 3.0 m/s collides inelastically with a stationary 10.0 kg mass. The two stick together and move off at the same speed. a.) What is the final speed? b.) How much kinetic energy is lost? 2 m/s 30 J 62.) A railroad car with a mass of 5.0 x 10 5 kg collides with a stationary railroad car of equal mass. After the collision, the two cars lock together and move off at 4.0 m/s. a.) Before the collision, the first railroad car moved at 8.0 m/s. What was its momentum? b.) What is the total momentum of the two cars after the collision? c.) Find the kinetic energies of the two cars before and after the collision. d.) Account for the loss of total kinetic energy in the system? 4.0 x 10 6 N s 1.6 x 10 7 J Energy Considerations of a Pendulum: The swinging of a pendulum illustrates the way in which potential energy may be transformed into kinetic energy and then back to potential energy. When a pendulum is pulled to one side, the mass has no velocity and no kinetic energy, but it does have potential energy. When the mass is released, it swings downward, acquiring kinetic energy and losing potential energy. At the bottom of the swing, the pendulum bob has maximum kinetic energy. As it swings upward, this kinetic energy is transformed again into potential energy. A very similar transformation of energy can be traced in the case of a mass on the end of a spring also. 63.) A pendulum bob is pulled to one side until its center of gravity has been raised 10.0 cm above its equilibrium position. Find the speed of the bob as it swings through the equilibrium position. 1.4 m/s 64.) A pendulum bob has a mass of 0.5 kg. It is suspended by a cord 2.0 m long, which is pulled back through an angle of 30. Find its maximum potential energy relative to its lowest position and its potential energy when the cord makes an angle of 15 with the vertical. Find its maximum speed and its speed when the cord makes the angle of 15 with the vertical. 1.31 J 0.33 J 2.29 m/s 1.98 m/s The ballistic pendulum is a device, which makes possible the determination of the velocity of a projectile. The projectile is fired into a block supported as a simple pendulum. The block swings up and is caught at its maximum height. Its change in potential energy is calculated. From the change in potential energy, the change in kinetic energy is known. From the change in kinetic energy, the velocity after the collision can be calculated. Then use the conservation of momentum to calculate the initial velocity of the projectile. 65.) A bullet of 20.0 g mass is fired into a block of 500.0 g mass supported from a 100.0 cm cord, causing it to swing through an angle of 10 with the vertical. Calculate the original velocity of the bullet. 14 m/s 66.) A bullet fired at a ballistic pendulum lodges in the wooden block which swings over and rises to a height of 7.0 cm. The mass of the bullet is 6.0 g and the mass of the block is 4.0 kg. The pendulum string is 5.0 m long. What is the original velocity of the bullet? 782 m/s

More About Potential Energy: As discussed earlier, objects that have been raised short distances above the earth have energy stored in them. This energy (due to an object s position above the earth s surface) may be used to do work on the object s surroundings should the object fall back to earth. Objects may also have the ability to do work because of their position relative to forces produced by other objects that are pushing or pulling on them. For example: the spring of a pinball machine has energy because it is compressed, or the twine (of a bow and arrow) because it has been drawn back ready to fire. This potential energy is called Elastic Potential Energy and shall be indicated as U elas or simply U e. For a spring, the amount of Ue depends directly on the spring constant and is given as: U e = ½ kx 2 67.) A 0.30 kg block sliding on a horizontal frictionless surface with a speed of 2.5 m/s strikes a light spring that has a spring constant of 3.0 x 10 3 N/m (see the diagram below). a.) What is the total mechanical energy of the system? b.) What is the kinetic energy of the block when the spring is compress a distance of 1.0 cm? (Assuming that no energy is lost in collision) c.) How far will the spring be compressed when the block comes to a stop? a.) 0.94 J b.) 0.79 J v m 68.) A 0.25 kg marble is placed on a vertical spring (spring constant 10 N/m) which compresses the spring a little bit. The marble sitting on the spring is then compressed an additional 50 cm and then released make sure you re out of the way! a.) How much elastic potential energy is contained in the system at full compression? b.) Assuming no energy loss, how high does the marble rise above the spring? c.) If the marble falls back down directly onto the spring, by how much does the spring get compressed? 2.78 J 1.13 m 74.5 cm