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**Physics C Energy Thursday, Jan 6, 2011! Happy New Year.**

3/25/2017 Introduction to Work Thursday, Jan 6, 2011! Happy New Year. Bertrand

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Physics C Energy 3/25/2017 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. Bertrand

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**Calculating Work – Step 1, constant F**

Physics C Energy 3/25/2017 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 Bertrand

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**Forces can do positive or negative work.**

Physics C Energy 3/25/2017 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 Bertrand

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**Units of Work SI System: Joule (N m) Physics C Energy 3/25/2017**

Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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**The work done by the rope on the sled**

Physics C Energy 3/25/2017 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. Bertrand

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**Work and a Pulley System**

Physics C Energy 3/25/2017 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. Bertrand

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Physics C Energy 3/25/2017 Work as a “Dot Product” Bertrand

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**Calculating Work a Different Way**

Physics C Energy 3/25/2017 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 Bertrand

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**The “scalar product” of two vectors is called the “dot product”**

Physics C Energy 3/25/2017 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. Bertrand

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**Why is work a dot product?**

Physics C Energy 3/25/2017 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. Bertrand

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Physics C Energy 3/25/2017 Problem: Vector A has a magnitude of 8.0 and vector B has a magnitude of The two vectors make an angle of 40o with each other. Find A•B. Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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**Homework 1/6/11 Page 160-163: 18, 19, 21, 51, 56 Physics C Energy**

3/25/2017 Homework 1/6/11 Page : 18, 19, 21, 51, 56 Bertrand

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**Work by Variable Forces**

Physics C Energy 3/25/2017 Work by Variable Forces Sometimes, forces are not constant. Examples, spring forces, air resistant forces. How do we calculate work in these situations? Bertrand

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**Work and Variable Forces**

Physics C Energy 3/25/2017 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 Bertrand

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**Work and variable force**

Physics C Energy 3/25/2017 Work and variable force The area under the curve of a graph of force vs displacement gives the work done by the force. F(x) x xa xb W = F(x) dx xa xb Bertrand

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Physics C Energy Problem: Determine the work done by the force as the particle moves from x = 2 m to x = 8 m. 3/25/2017 We already know this technique manually with a graph; we can now call this ‘manual integration’ F (N) 20 40 -20 -40 2 4 6 8 10 12 x (m) Bertrand

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**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”.

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**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.

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**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, xf and 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.

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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.)

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**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.)

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**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.)

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**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.)

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**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.)

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**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.)

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**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?

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Physics C Energy 3/25/2017 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. Bertrand

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**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?

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**Try these samples on a clean sheet of paper.**

Physics C Energy 3/25/2017 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. Bertrand

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Physics C Energy 3/25/2017 Problem: Derive an expression for the work done by a spring as it is stretched from its equilibrium position Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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Physics C Energy 3/25/2017 Work Energy Theorem Bertrand

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Physics C Energy 3/25/2017 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 Bertrand

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**The Work-Energy Theorem**

Physics C Energy 3/25/2017 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.) Bertrand

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Physics C Energy 3/25/2017 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 Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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Physics C Energy 3/25/2017 Problem: Calculate the kinetic energy change of a 3.0 kg object that changes its velocity from (2.0 i j -1.0 k) m/s to (-1.0 i j -2.0 k) m/s. How much net work done on this object? Bertrand

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Physics C Energy 3/25/2017 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? Bertrand

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**Power Power is the rate of which work is done.**

Physics C Energy 3/25/2017 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 Bertrand

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**Units of Power Watt = J/s ft lb / s horsepower 550 ft lb / s 746 Watts**

Physics C Energy 3/25/2017 Units of Power Watt = J/s ft lb / s horsepower 550 ft lb / s 746 Watts Bertrand

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Physics C Energy 3/25/2017 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. Bertrand

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Physics C Energy 3/25/2017 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%? Bertrand

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Physics C Energy 3/25/2017 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. Bertrand

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Physics C Energy 3/25/2017 Problem: Using what you know about units, calculate how many Joules is in a kilowatt-hour. Bertrand

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**Conservative and Non-Conservative Forces**

Physics C Energy 3/25/2017 Conservative and Non-Conservative Forces Bertrand

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**More about force types Conservative forces: Non-conservative forces:**

Physics C Energy 3/25/2017 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 Bertrand

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Physics C Energy 3/25/2017 Potential Energy, U A type of mechanical energy possessed by an object by virtue of its position or configuration. Represented by the letter U. Examples: Gravitational potential energy, Ug. Electrical potential energy , Ue. Spring potential energy , Us. The work done by conservative forces is the negative of the potential energy change. W = -ΔU Bertrand

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**Gravitational Potential Energy (Ug)**

Physics C Energy 3/25/2017 Gravitational Potential Energy (Ug) The change in gravitational potential energy is the negative of the work done by gravitational force on an object when it is moved. For objects near the earth’s surface, the gravitational pull of the earth is roughly constant, so the force necessary to lift an object at constant velocity is equal to the weight, so we can say ΔUg = -Wg = mgh Note that this means we have defined the point at which Ug = 0, which we can do arbitrarily in any given problem close to the earth’s surface. h Fapp mg Bertrand

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**Spring Potential Energy, Us**

Physics C Energy 3/25/2017 Spring Potential Energy, Us Springs obey Hooke’s Law. Fs(x) = -kx Fs is restoring force exerted BY the spring. Ws = Fs(x)dx = -k xdx Ws is the work done BY the spring. Us = ½ k x2 Unlike gravitational potential energy, we know where the zero potential energy point is for a spring. Bertrand

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Physics C Energy 3/25/2017 X Y Z Problem: Three identical springs (X, Y, and Z) are hung as shown. When a 5.0-kg mass is hung on X, the mass descends 4.0 cm from its initial point. When a 7.0-kg mass is hung on Z, how far does the mass descend? Bertrand

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**Hint 2: ΔU = -Wc (and gravity is conservative!) **

Physics C Energy 3/25/2017 Sample problem: Gravitational potential energy for a body a large distance r from the center of the earth is defined as shown below. Derive this equation from the Universal Law of Gravity. Hint 1: dW = F(r)•dr Hint 2: ΔU = -Wc (and gravity is conservative!) Hint 3: Ug is zero at infinite separation of the masses. Bertrand

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**Conservation of Mechanical Energy**

Physics C Energy 3/25/2017 Conservation of Mechanical Energy Bertrand

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Physics C Energy 3/25/2017 System Boundary Environment System Bertrand

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**Law of Conservation of Energy**

Physics C Energy 3/25/2017 Law of Conservation of Energy The system is isolated and boundary allows no exchange with the environment. E = U + K + Eint = Constant No mass can enter or leave! No energy can enter or leave! Energy is constant, or conserved! Bertrand

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**Law of Conservation of Mechanical Energy**

Physics C Energy 3/25/2017 Law of Conservation of Mechanical Energy E = U + K = Constant We only allow U and K to interchange. We ignore Eint (thermal energy) Bertrand

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**Law of Conservation of Mechanical Energy**

Physics C Energy 3/25/2017 Law of Conservation of Mechanical Energy E = U + K = C or E = U + K = 0 for gravity Ug = mghf - mghi K = ½ mvf2 - ½ mvi2 (What assumptions are we making here?) for springs Us = ½ kxf2 - ½ kxi2 Ranking Tasks 1 Bertrand

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**Pendulum Energy h ½mv12 + mgh1 = ½mv22 + mgh2**

Physics C Energy 3/25/2017 Pendulum Energy h ½mv12 + mgh1 = ½mv22 + mgh2 For any points two points in the pendulum’s swing Bertrand

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**Spring Energy m -x m m x ½ kx12 + ½ mv12 = ½ kx22 + ½ mv22**

Physics C Energy 3/25/2017 Spring Energy m -x ½ kx12 + ½ mv12 = ½ kx22 + ½ mv22 For any two points in a spring’s oscillation m m x Bertrand

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**What is the change in potential energy of the system?**

Physics C Energy 3/25/2017 Problem: A single conservative force of F = (3i + 5j) N acts on a 4.0 kg particle. Calculate the work done if the particle if the moves from the origin to r = (2i - 3j) m. Does the result depend on path? What is the speed of the particle at r if the speed at the origin was 4.0 m/s? What is the change in potential energy of the system? Bertrand

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**Sample Problem: A bead slides on the loop-the-loop shown**

Sample Problem: A bead slides on the loop-the-loop shown. If it is released from height h = 3.5 R, what is the speed at point A? How great is the normal force at A if the mass is 5.0 g? Physics C Energy 3/25/2017 Bertrand

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**Non-conservative Forces and Conservation of Energy**

Physics C Energy 3/25/2017 Non-conservative Forces and Conservation of Energy Bertrand

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**Non-conservative forces**

Physics C Energy 3/25/2017 Non-conservative forces Non-conservative forces change the mechanical energy of a system. Examples: friction and drag Wtot = Wnc + Wc = DK Wnc = DK – Wc Wnc = DK + DU Bertrand

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Physics C Energy 3/25/2017 Sample Problem: A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with the horizontal. If an average friction force of 4,000 N impedes the motion of the car, find the speed of the car at the bottom of the driveway. Bertrand

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Physics C Energy 3/25/2017 Problem: A parachutist of mass 50 kg jumps out of a hot air balloon 1,000 meters above the ground and lands on the ground with a speed of 5.00 m/s. How much energy was lost to friction during the descent? Bertrand

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**Force and Potential Energy**

Physics C Energy 3/25/2017 Force and Potential Energy In order to discuss the relationships between potential energy and force, we need to review a couple of relationships. Wc = FDx (if force is constant) Wc = Fdx = - dU = -DU (if force varies) Fdx = - dU Fdx = -dU F = -dU/dx Bertrand

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**Stable Equilibrium – 1st and 2nd Derivatives**

Physics C Energy 3/25/2017 Stable Equilibrium – 1st and 2nd Derivatives U x Bertrand

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**Unstable Equilibrium – 1st and 2nd Derivatives**

Physics C Energy 3/25/2017 Unstable Equilibrium – 1st and 2nd Derivatives U x Bertrand

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**Neutral Equilibrium – 1st and 2nd Derivatives**

Physics C Energy 3/25/2017 Neutral Equilibrium – 1st and 2nd Derivatives U x Bertrand

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**More on Potential Energy and Force**

Physics C Energy 3/25/2017 More on Potential Energy and Force In multiple dimensions, you can take derivatives of each dimension separately. F(r) = -dU(r)/dr Fx = -U/ x Fy = -U/ y Fz = -U/ z F = Fxi + Fyj + Fzk Bertrand

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**Molecular potential energy diagrams**

Physics C Energy 3/25/2017 Molecular potential energy diagrams U R Bertrand

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Physics C Energy 3/25/2017 Problem: The potential energy of a two-particle system separated by a distance r is given by U(r) = A/r, where A is a constant. Find the radial force F that each particle exerts on the other. Bertrand

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Physics C Energy 3/25/2017 Problem: A potential energy function for a two-dimensional force is of the form U = 3x3y – 7x. Find the force acting at a point (x,y). Bertrand

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**Physics C Energy October 31, 2008**

3/25/2017 Linear Momentum October 31, 2008 Bertrand

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**Announcements Turn in homework due today: Chapter 8, problems 28,29,31**

Physics C Energy 3/25/2017 Announcements Turn in homework due today: Chapter 8, problems 28,29,31 Next week, W-F, Rocket Project Bertrand

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Physics C Energy 3/25/2017 Linear Momentum Momentum is a measure of how hard it is to stop or turn a moving object. p = mv (single particle) P = Σpi (system of particles) Bertrand

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**Problem: How fast must an electron move to have the same momentum as a proton moving at 300 m/s?**

Physics C Energy 3/25/2017 Bertrand

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**Problem: A 90-kg tackle runs north at 5**

Problem: A 90-kg tackle runs north at 5.0 m/s and a 75-kg quarterback runs east at 8.0 m/s. What is the momentum of the system of football players? Physics C Energy 3/25/2017 Bertrand

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**Impulse (J) Impulse is the integral of force over a period of time.**

Physics C Energy 3/25/2017 Impulse (J) Impulse is the integral of force over a period of time. J = Fdt The change in momentum of a particle is equal to the impulse acting on it. Dp = J Bertrand

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**Problem: Restate Newton’s 2nd Law in terms of impulse.**

Physics C Energy 3/25/2017 Problem: Restate Newton’s 2nd Law in terms of impulse. Bertrand

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Physics C Energy 3/25/2017 Force Impulsive forces are generally of high magnitude and short duration. time Bertrand

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**If the collision took 2.3 ms, what was the average force?**

Physics C Energy 3/25/2017 Problem: A 150-g baseball moving at 40 m/s 15o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35o above the horizontal. What is the impulse exerted by the bat on the ball? If the collision took 2.3 ms, what was the average force? Bertrand

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Physics C Energy 3/25/2017 Problem: An 85-kg lumberjack stands at one end of a floating 400-kg log that is at rest relative to the shore of a lake. If the lumberjack jogs to the other end of the log at 2.5 m/s relative to the shore, what happens to the log while he is moving? Bertrand

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Physics C Energy 3/25/2017 Problem: Two blocks of mass 0.5 kg and 1.5 kg are placed on a horizontal, frictionless surface. A light spring is compressed between them. A cord initially holding the blocks together is burned; after this, the block of mass 1.5 kg moves to the right with a speed of 2.0 m/s. A) What is the speed and direction of the other block? B) What was the original elastic energy in the spring? Bertrand

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**Conservation of Linear Momentum**

Physics C Energy 3/25/2017 Conservation of Linear Momentum The linear momentum of a system is conserved unless the system experiences an external force. Bertrand

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**Physics C Energy Collisions in 1 and 2 Dimension**

3/25/2017 Collision Review Collisions in 1 and 2 Dimension Bertrand

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**Collision review In all collisions, momentum is conserved.**

Physics C Energy 3/25/2017 Collision review In all collisions, momentum is conserved. Elastic Collisions: No deformation occurs Kinetic energy is also conserved. Inelastic Collisions: Deformation occurs Kinetic energy is lost. Perfectly Inelastic Collisions Objects stick together, kinetic energy is lost. Explosions Reverse of perfectly inelastic collision, kinetic energy is gained. Bertrand

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Physics C Energy 3/25/2017 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides with a stationary 0.5 kg cart and sticks to it. At what speed are the carts moving after the collision? Bertrand

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Physics C Energy 3/25/2017 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides elastically with a stationary 0.5 kg cart. At what speed are each of the carts moving after the collision? Bertrand

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Physics C Energy 3/25/2017 1 Dimensional Problem: What is the recoil velocity of a 120-kg cannon that fires a 30-kg cannonball at 320 m/s? Bertrand

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**For 2-dimensional collisions**

Physics C Energy 3/25/2017 For 2-dimensional collisions Use conservation of momentum independently for x and y dimensions. You must resolve your momentum vectors into x and y components when working the problem Bertrand

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2 Dimensional Problem: A pool player hits a cue ball in the x-direction at 0.80 m/s. The cue ball knocks into the 8-ball, which moves at a speed of 0.30 m/s at an angle of 35o angle above the x-axis. Determine the angle of deflection of the cue ball. Physics C Energy 3/25/2017 Bertrand

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**Collision Problems Practice Day**

Physics C Energy 3/25/2017 Collision Problems Practice Day Bertrand

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Physics C Energy 3/25/2017 Problem 1 A proton, moving with a velocity of vii, collides elastically with another proton initially at rest. If the two protons have equal speeds after the collision, find a) the speed of each in terms of vi and b) the direction of the velocity of each. Bertrand

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Physics C Energy 3/25/2017 Center of Mass Bertrand

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Physics C Energy 3/25/2017 Center of Mass The center of mass of a system is the point at which all the mass can be assumed to reside. Sometimes the system is an assortment of particles and sometimes it is a solid object. Mathematically, you can think of the center of mass as a “weighted average”. Bertrand

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**Center of Mass – system of points**

Physics C Energy 3/25/2017 Center of Mass – system of points Analogous equations exist for velocity and acceleration, for example Bertrand

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**Sample Problem y 2 x -2 -2 2 4 Determine the Center of Mass. 2 kg 3 kg**

Physics C Energy 3/25/2017 Determine the Center of Mass. Sample Problem y 2 2 kg x 3 kg 1 kg -2 -2 2 4 Bertrand

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**Center of Mass -- solid objects**

Physics C Energy 3/25/2017 x Center of Mass -- solid objects If the object is of uniform density, you pick the geometric center of the object. x x x For combination objects, pick the center of each part and then treat each center as a point. Bertrand

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**Center of Mass Problem y Determine the center of mass of this shape. x**

Physics C Energy 3/25/2017 Center of Mass Problem x y 2R Determine the center of mass of this shape. Bertrand

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**Center of Mass for more complicated situations**

Physics C Energy 3/25/2017 Center of Mass for more complicated situations If the shapes are points or simple geometric shapes of constant density, then Anything else is more complicated, and Bertrand

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Problem: Find the x-coordinate of the center of mass of a rod of length L whose mass per unit length varies according to the expression l = ax. Physics C Energy 3/25/2017 L x y Bertrand

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Problem: A thin strip of material of mass M is bent into a semicircle of radius R. Find its center of mass. Physics C Energy 3/25/2017 x y R -R Bertrand

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**Motion of a System of Particles**

Physics C Energy 3/25/2017 Motion of a System of Particles Bertrand

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**Motion of a System of Particles**

Physics C Energy 3/25/2017 Motion of a System of Particles If you have a problem involving a system of many particles, you can often simplify your problem greatly by just considering the motion of the center of mass. If all forces in the sytem are internal, the motion of the center of mass will not change. The effect of an external force that acts on all particles can be simplified by considering that it acts on the center of mass; that is: SFext = Macm Bertrand

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**Homework problem – collaborative work (15 minutes)**

Physics C Energy 3/25/2017 Homework problem – collaborative work (15 minutes) This problem is a toughie! However, it gets easier if you use the idea that the center of mass of the system of Romeo, Juliet, and canoe does not change because all forces are internal. The question is paraphrased below: Romeo (77 kg) sits in the back of a canoe 2.70 m away from Juliet (55 kg) who sits in the front. The canoe has a mass of 80 kg. Juliet moves to the rear of the canoe to kiss Romeo. How far does the canoe move forward when she does this? (Assume a canoe that is symmetrical). Bertrand

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