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## Introduction to Work

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**Introduction to Work**Thursday, Jan 6, 2011! Happy New Year.**Energy and Work**• A body experiences a change in energy when one or more forces do work on it. A body must move under the influence of a force or forces to say work was done. • A force does positive work on a body when the force and the displacement are at least partially aligned. Maximum positive work is done when a force and a displacement are in exactly the same direction. • If a force causes no displacement, it does zero work. • Forces can do negative work if they are pointed opposite the direction of the displacement.**Calculating Work – Step 1, constant F**• If a force on an object is at least partially aligned with the displacement of the object, positive work is done by the force. The amount of work done depends on the magnitude of the force, the magnitude of the displacement, and the degree of alignment. • W= F r cos q F F r**Forces can do positive or negative work.**• When the load goes up, gravity does negative work and the crane does positive work. • When the load goes down, gravity does positive work and the crane does negative work. • Ranking Task 1 F mg**Units of Work**• SI System: • Joule (N m)**Problem: A droplet of water of mass 50 mg falls at constant**speed under the influence of gravity and air resistance. After the drop has fallen 1.0 km, what is the work done by a) gravity and b) air resistance?**Problem: A sled loaded with bricks has a mass of 20.0 kg. It**is pulled at constant speed by a rope inclined at 25o above the horizontal, and it moves a distance of 100 m on a horizontal surface. If the coefficient of kinetic friction between the sled and the ground is 0.40, calculate • The tension in the rope. • The work done by the rope on the sled • The work done by friction on the sled.**F**m Work and a Pulley System • A pulley system, which has at least one pulley attached to the load, can be used to reduce the force necessary to lift a load. • Amount of work done in lifting the load is not changed. • The distance the force is applied over is increased, thus the force is reduced, since W = Fd.**Calculating Work a Different Way**• Work is a scalar resulting from the multiplication of two vectors. • We say work is the “dot product” of force and displacement. • W = F • r • dot product representation • W= F r cos q • useful if given magnitudes and directions of vectors • W = Fxrx + Fyry + Fzrz • useful if given unit vectors**The “scalar product” of two vectors is called the “dot**product” • The “dot product” is one way to multiply two vectors. (The other way is called the “cross product”.) • Applications of the dot product • Work W = F d • Power P = F v • Magnetic Flux ΦB = B A • The quantities shown above are biggest when the vectors are completely aligned and there is a zero angle between them.**Why is work a dot product?**F s W = F • r W = F r cos Only the component of force aligned with displacement does work.**Problem: Vector A has a magnitude of 8.0 and vector B has a**magnitude of 12.0. The two vectors make an angle of 40o with each other. Find A•B.**Problem: A force F = (5.0i + 6.0j – 2.0k)N acts on an**object that undergoes a displacement of r = (4.0i – 9.0j + 3.0k)m. How much work was done on the object by the force?**Problem: A force F = (5.0i – 3.0j) N acts upon a body**which undergoes a displacement d = (2.0i – j) m. How much work is performed, and what is the angle between the vectors?**Homework 1/6/11**• Page 160-163: 18, 19, 21, 51, 56**Work by Variable Forces**Sometimes, forces are not constant. Examples, spring forces, air resistant forces. How do we calculate work in these situations?**Work and Variable Forces**• For constant forces • W = F • r • For variable forces, you can’t move far until the force changes. The force is only constant over an infinitesimal displacement. • dW = F • dr (dr, dW = small, small sample where force could be considered constant) • To calculate work for a larger displacement, you have to take an integral • W = dW = F • dr**F(x)**xb W = F(x) dx xa x xa xb Work and variable force The area under the curve of a graph of force vs displacement gives the work done by the force.**F (N)**40 20 0 2 4 6 8 10 12 x (m) -20 -40 We already know this technique manually with a graph; we can now call this ‘manual integration’ • Problem: Determine the work done by the force as the particle moves from x = 2 m to x = 8 m.**Tutorial on the Integral**• Mathematically, the integral is used to calculate a sum composed of many, many tiny parts. • In physics, the integral is used to find a measurable change resulting from very small incremental changes. It’s useful to think of it as a fancy addition process or a multiplication process, depending on the situation. • Here’s a specific example. Velocity times time gives displacement. If the velocity is changing with time, but a very tiny time change is used to calculate a very tiny displacement, we can nonetheless assume the velocity was constant during that tiny time change. • If we calculate tiny displacements this way (recalculating our velocity for each time increment), then add the tiny displacements up to get a larger displacement, we have done “integration”.**Velocity as a Function of Time Displacement**• Velocity can be represented as follows: • Rearrangement of this expression yields: • What this means is that we can calculate a tiny displacement dx from the velocity v at a given time times a tiny time increment dt.**Summing the Displacements**• When we sum up these tiny displacements, we use the following notation: • This notation indicates that we are summing up all the little displacements dx starting at position xo at time to until we reach a final position and time, xfand tf. The velocity v may be a function of time, and may be slightly different for one time increment dt and the next time increment.**Evaluating Integrals**• We will evaluate polynomial integrals by reversing the process we used in taking a polynomial derivative. This process is sometimes called “anti-differentiation”, or “doing an anti-derivative”. • The general method for doing an anti-derivative is: • Can you see how this is the reverse of taking a derivative? (Note: This “indefinite integral” requires us to add a constant C to compensate for constants that may have been lost during differentiation…but we will do “definite integrals” that do not require C.)**Evaluating Definite Integrals**• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)**Evaluating Definite Integrals**• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)**Evaluating Definite Integrals**• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)**Evaluating Definite Integrals**• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)**Evaluating Definite Integrals**• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)**Sample problem: Consider a force that is a function of time:**F(t) = (3.0 t – 0.5 t2)N • If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle?**Problem: A force acting on a particle is Fx = (4x – x2)N.**Find the work done by the force on the particle when the particle moves along the x-axis from x= 0 to x = 2.0 m.**Sample problem: Consider a force that is a function of time:**F(t) = (16 t2 – 8 t + 4)N • If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle?**Try these samples on a clean sheet of paper.**• 1. Calculate the work done by a force that applies F(x) = 3x + x3, while it moves an object from 3 to 8 meters. • 2. A tractor pulls a trailer with a force that varies according to F(x) = x – 3x. Calculate the work done from rest to 2.0 m. • 3.**Problem: Derive an expression for the work done by a spring**as it is stretched from its equilibrium position**Problem: How much work does an applied force do when it**stretches a nonlinear spring where the force varies according to the expressions F = (300 N/m) x – (25 N/m2)x2 from its equilibrium length to 20 cm?**Net Work or Total Work**• An object can be subject to many forces at the same time, and if the object is moving, the work done by each force can be individually determined. • At the same time one force does positive work on the object, another force may be doing negative work, and yet another force may be doing no work at all. • The net work, or total, work done on the object (Wnet or Wtot) is the scalar sum of the work done on an object by all forces acting upon the object. • Wnet = ΣWi**The Work-Energy Theorem**• Wnet = ΔK • When net work due to all forces acting upon an object is positive, the kinetic energy of the object will increase. • When net work due to all forces acting upon an object is negative, the kinetic energy of the object will decrease. • When there is no net work acting upon an object, the kinetic energy of the object will be unchanged. • (Note this says nothing about the kinetic energy.)**Kinetic Energy**• Kinetic energy is one form of mechanical energy, which is energy we can easily see and characterize. Kinetic energy is due to the motion of an object. • K = ½ m v2 • K: Kinetic Energy in Joules. • m: mass in kg • v: speed in m/s • In vector form, K = ½ m v•v • Ranking Tasks 1234**Problem: A net force of 320 N acts over 1.3 m on a 0.4 kg**particle moving at 2.0 m/s. What is the speed of this particle after this interaction?**Problem: Calculate the kinetic energy change of a 3.0 kg**object that changes its velocity from (2.0 i + 2.0 j -1.0 k) m/s to (-1.0 i + 1.0 j -2.0 k) m/s. How much net work done on this object?**Problem: A force of F1 = (4.0 i + j) N and another of F2 =**-4.0 j N act upon a 1 kg object at rest at the origin. What is the speed of the object after it has moved a distance of 3.0 m?**Power**• Power is the rate of which work is done. • No matter how fast we get up the stairs, our work is the same. • When we run upstairs, power demands on our body are high. • When we walk upstairs, power demands on our body are lower. • Pave = W / t • Pinst = dW/dt • P = F • v**Units of Power**• Watt = J/s • ft lb / s • horsepower • 550 ft lb / s • 746 Watts**Problem: A 1000-kg space probe lifts straight upward off the**planet Zombie, which is without an atmosphere, at a constant speed of 3.0 m/s. What is the power expended by the probe’s engines? The acceleration due to gravity of Zombie is ½ that of earth’s.**Problem: Develop an expression for the power output of an**airplane cruising at constant speed v in level flight. Assume that the aerodynamic drag force is given by FD = bv2. • By what factor must the power be increased to increase airspeed by 25%?**How We Buy Energy…**• The kilowatt-hour is a commonly used unit by the electrical power company. • Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed.**Problem: Using what you know about units, calculate how many**Joules is in a kilowatt-hour.**More about force types**• Conservative forces: • Work in moving an object is path independent. • Work in moving an object along a closed path is zero. • Work is directly related to a negative change in potential energy • Ex: gravity, electrostatic, magnetostatic, springs • Non-conservative forces: • Work is path dependent. • Work along a closed path is NOT zero. • Work may be related to a change in mechanical energy, or thermal energy • Ex: friction, drag, magnetodynamic