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Work and Energy

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Product of Vectors

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Product of Vectors We know, how to add and subtract vectors. Vectors are not ordinary numbers, so ordinary multiplication is not directly applicable to vectors. We will define two kinds of product of vectors: Scalar product: yields a result that is a scalar quantity. Vector product: yields another vector.

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Scalar Product Scalar product of two vectors is a scalar quantity. Scalar product can be negative, positive or zero. It can be expressed in two ways: In terms of magnitudes of A and B and the angle between them: In terms of components of A and B: ranges from 0° to 180°

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**Scalar Product Positive, when is between 0° to 90°**

Scalar product is: Positive, when is between 0° to 90° Negative, when is between 90° to 180° Zero? The scalar product of two perpendicular vectors is always zero. Scalar product obeys the communicative law of multiplication: the order of the two vectors does not matter. Unit vectors are to each other. Scalar product of unit vectors:

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Work

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**Work Push a stalled car off the road**

Everyday life concept of work: any activity that requires muscular or mental effort, for example Push a stalled car off the road Complete MasteringPhysics homework in time Physics concept of work: The total work done on a particle by all forces that act on it equals the change in its kinetic energy (quantity that is related to the particle’s speed even non-constant speed and non-constant forces; we will learn this later) The work done by a constant force on an object that is undergoing a straight line displacement is given by

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**Work You do more work if the force is greater**

The work done by a constant force on an object that is undergoing a straight line displacement is given by Definition of work is based on observations. You do work by exerting the force on an object while that object moves from one place to another (undergoes displacement): You do more work if the force is greater You do more work if displacement is greater

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**Work Unit of force is pound, unit of distance is foot**

James Joule SI unit of work is Joule (J) “Jewel” 1 Joule = (1 Newton) (1 meter); 1 J = 1 N·m British unit of work is foot-pound (ft·lb) Unit of force is pound, unit of distance is foot Conversion: 1 J = ft·lb, 1 ft·lb = J

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Work You push a stalled car through a displacement S with a constant force F in the direction of motion: You push a stalled car through a displacement S with a constant force F at angle to the direction of motion: Only component of force in direction of car’s displacement is important

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**Work and Kinetic Energy**

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**Work and Kinetic Energy**

From the definition of work we know that the total work done on an object is related to its displacement (changes in position). Work is also related to changes in the speed of the object.

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**Work and Kinetic Energy**

Example: Block sliding on a frictionless table Forces acting on a block: its weight, normal force, and the force F exerted by the hand.

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**Work and Kinetic Energy**

Example: Block sliding on a frictionless table The net force on a block is in the direction of its motion. From N2L, this means that the block speeds up. W=F·S also tells us that the total work will be positive. Here only the component F·cos contributes to Wtotal. The block speeds up as well. Forces acting on a block: its weight, normal force, and the force F exerted by the hand.

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**Work and Kinetic Energy**

Example: Block sliding on a frictionless table The net force here opposes the displacement. From N2L, this means that the block slows down. W=F·S also tells us that the total work will be negative. Here the net force is zero, so the speed of the block stays the same, and Wtotal is zero. Forces acting on a block: its weight, normal force, and the force F exerted by the hand.

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**Work and Kinetic Energy**

When an object undergoes a displacement: object will "speed up" if Wtotal > 0, object will "slow down" if Wtotal < 0, object will "maintain the same speed“ (constant) if Wtotal = 0.

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**Work and Kinetic Energy**

Consider a particle with mass m moving along the x-axis under the action of a constant net force with magnitude F directed along the positive x-axis. Particle acceleration is constant, and by N2L: F=m·ax. Suppose, speed changes from v1 to v2 while particle undergoes a displacement S=x2-x1 from point x1 to x2. 1-D constant-acceleration equation:

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**Work and Kinetic Energy**

The product F·S is the work done by the net force. Thus, it is equal to the total work Wtotal done by all the forces acting on a particle. Definition of Kinetic Energy: Like work, kinetic energy of a particle is a scalar quantity: it depends on particle’s mass and speed, not its direction of motion. Car has the same kinetic energy when going north at 10m/s as when going east at 10m/s. Kinetic energy can never be negative; it’s zero when particle is at rest.

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Work - Energy Theorem Work done by the net force on a particle equals the change in the particle’s kinetic energy: Work - Energy Theorem: When an object moves: object will "speed up" if Wtotal > 0, K2 > K1 object will "slow down" if Wtotal < 0, K2 < K1 object will "maintain the same speed“ (constant) if Wtotal =0, K2=K1 Speeds and distances must be measured in inertial frame of reference! Kinetic energy and work have the same units (Joules, or N·m):

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**Work and Kinetic Energy**

Problem Solving Strategy IDENTIFY the relevant concepts: The work-energy theorem is extremely useful in situations where you want to relate a body’s speed at one point in its motion to its speed at a different point. This approach is less useful for problems that involve time, such as finding the time it takes a body to go from point 1 to point 2. The reason is that the work-energy theorem doesn’t involve time at all. For problems that involve time, it’s usually best to use the relationships among time, position, velocity, and acceleration we learned in 1-D motion. SET UP the problem using the following steps: Choose the initial and final positions of the body, and draw a free-body diagram showing all the forces that act on the body. Choose a coordinate system. (If the motion is along a straight line, it’s usually easiest to have both the initial and final positions lie along the x-axis.) List the unknown and known quantities, and decide which unknowns are your target variables. In some cases the target variable will be the body’s initial or final speed; in other cases it will be the magnitude of one of the forces acting on the body.

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**Work and Kinetic Energy**

Problem Solving Strategy EXECUTE: Calculate the work done by each force. Be sure to check the sign of the work for each force; it must be positive if the force has a component in the direction of the displacement, negative if it has a component opposite the displacement, and zero if the force and displacement are perpendicular. Add the amounts of work done by each force to find the total work. Be careful with signs! Sometimes it may be easier to calculate the vector sum of the forces (the net force), then find the work done by the net force. Write expressions for the initial and final kinetic energies, K1 and K2. Note that kinetic energy involves mass, not weight; if you are given the body’s weight, you’ll need to find the mass. Finally, use the relationship Wtot=K2-K1 to solve for the target variable. Remember that the right-hand side of this equation is the final kinetic energy minus the initial kinetic energy, never the other way around. EVALUATE your answer: Check whether your answer makes physical sense. A key item to remember is that kinetic energy can never be negative. If you come up with a negative value of K, you’ve made a mistake!

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Kinetic Energy The example with the hammerhead gives insight into the physical meaning of kinetic energy. The hammerhead was dropped from rest, and its kinetic energy when it hits the I-beam equals the total work done on it up to that point by the net force. To accelerate a particle with mass m from rest (zero kinetic energy) up to a speed v, the total work done on it must equal the change in kinetic energy from zero to K=0.5·m·v2: Kinetic energy of a particle is equal to the total work that was done to accelerate it from rest to its present speed. Or from its present speed to rest! Catch the ball right pull your hand back, increasing distance to stop the ball: ball does the work on your hand equal to the ball’s initial kinetic energy Wtot=FS=0.5mv2. Pulling hand back, you maximize the distance over which this force acts and thus minimize the force on your hand.

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Composite Systems Man standing on frictionless roller skates on a level surface, pushes against the rigid wall, setting himself in motion to the right. Forces acting on him: his weight W, upward normal forces n1 and n2 exerted by the ground on his skates, and the horizontal force F exerted on him by the wall. No vertical displacement, so forces W, n1 and n2 do NO work. Force F exerted on him by the wall is horizontal force that accelerates him to the right, but his hands don’t move. So this force doesn’t do work as well. Where does the man’s kinetic energy come from? This system is not a single point (or particle) Various parts of the body interact with each other (here: his hands are still, but his torso moves) Total kinetic energy of this composite system of body parts can change, even no work is done by forces applied outside the system.

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**Work and Energy with Varying Forces**

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**Work and Energy with Varying Forces**

We know that work done by a constant force on an object that is undergoing a straight line displacement is given by What happens when force exerted on an object is NOT constant and the object moves in path which is NOT straight? Example: spring, stretched More you stretch it, the harder you have to pull: thus the force is non-constant Let’s consider straight-line motion with non-constant force One complication at a time! Fx change along the x-axis (force depends on position)

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**Work and Energy with Varying Forces**

Particle moves from x1 to x2; Fx depends on coordinate x. Let’s divide the total displacement by small segments xa, xb, xc… Total work done during segment xa: ~ by the average force Fa in this segment multiplied by the displacement xa. All segments:

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**Work and Energy with Varying Forces**

If number of segments is very large, segment's width x is very small In the limit, the sum is integral of Fx from x1 to x2. Varying x-component of force, straight-line displacement On a graph of force as function of position, the total work done by this force is represented by the area under the curve between the initial and final positions.

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**Work and Energy with Varying Forces**

Let’s check it: if Fx is constant from x1 to x2.:

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Hooke’s Law To keep ideal spring stretched by amount x beyond its initial length, we need to apply on the end the force which is proportional to x: Force required to stretch a spring K – spring (force) constant, [N/m] Example: Force constant k Floppy toy spring: k=1 N/m Car suspension spring: k=105 N/m Robert Hooke 1635 – 1703 ... lean, bent and ugly man ...

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**Hooke’s Law What happens if you compress the spring?**

To stretch a spring, you must do work Suppose one end of a spring is fixed, you apply force on another end That end moves, so the force does work Work done by the force when spring elongation goes from zero to X: Total work is ~ to square of final elongation Graph, Area under the curve: If spring was initially already stretched a distance x1, the work to stretch it to a greater elongation x2: What happens if you compress the spring?

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**Hooke’s Law What happens if you compress the spring?**

Compression: Force Fx and displacement x are both negative Force is in the same direction as displacement: work is positive Example is following…

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**Varying Forces: Work - Energy Theorem**

One can use the same approach: divide total displacement into segments Apply Work-Energy Theorem for each segment: Wa=Ka=Faxa Sum the changes to find Wtot Another way: “Chain rule” for derivatives (dvx/dx)dx is the change in velocity dvx during displacement dx. Thus: Work – Energy Theorem is the same: valid for varying forces as well !

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**Curved Path: Work - Energy Theorem**

Force that varies in direction and magnitude Displacement lies along a curved path: particle moves from P1 to P2 Divide curve between P1 and P2 into small vector displacements d. Each d is a tangent to the path at its position. F is the force at a point along the path, is the angle between F and d. Small element of work W done on particle during displacement d : Total work done on a particle then: (Work done on a curved path)

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Power Watt’s Engine “Power man” Mr. Olympia

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**Power Definition of work makes no reference to the passage of time**

You lift weight 100N vertically at a distance 1m at constant velocity: You do (100N)(1m)=100J of work whether it takes 1 sec, 1 hour, 1 year… You want to know how quickly the work is done Power is the time rate at which work is done. Power is a scalar. Average power: Instantaneous power:

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**Power James Watt energy = power · time**

The SI unit of power is watt (W), 1 W = 1 Joule per 1 second. In the British system of units power is in ft·lb/sec or in a larger unit called horsepower (hp). 1 hp = 550 ft·lb/sec = 33,000 ft·lb/min = 746 W = ¾ of kilowatt (kW) The kilowatt-hour (kWh) is a unit of electrical energy, not power energy = power · time James Watt 1736 – 1819 Watt's steam engines …Started with nothing, died as a very wealthy man… In mechanics, power is expressed in terms of force and velocity In terms of scalar product

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James Watt Watt discovered how he could make the steam engine more efficient. Steam engine that cooled the used steam in a condenser separate from main cylinder. In 1755 Watt had been granted a patent by Parliament that prevented anybody else from making a steam-engine like the one he had developed. For next 25 years, the “Boulton & Watt” company had a virtual monopoly over the production of steam-engines. These machines were mainly sold to colliery owners who used them to pump water from their mines, and textile factories. Watt charged his customers a premium for using his steam engines. To justify this he compared his machine to a horse. Watt calculated that a horse exerted a pull of 180 lb., therefore, when he made a machine, he described its power in relation to a horse, i.e. "a 20 horse-power engine". Watt worked out how much each company saved by using his machine rather than a team of horses. The company then had to pay him one third of this figure every year, for the next twenty-five years. When James Watt died in 1819 he was a very wealthy man.

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Watt’s Steam Engine This Watt Rotation Engine is housed in the Henry Ford Museum in Dearborn, MI

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