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7.4 Work Done by a Varying Force

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Work Done by a Varying Force Assume that during a very small displacement, x, F is constant For that displacement, W ~ F x For all of the intervals,

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Work Done by a Varying Force, cont Sum approaches a definite value: Therefore:(7.7) The work done is equal to the area under the curve!! The work done is equal to the area under the curve!!

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Example 7.7 Total Work Done from a Graph (Example 7.4 Text Book) net work the area under the curve The net work done by this force is the area under the curve W = Area under the Curve W = A R + A T W = (B)(h) + (B)(h)/2 = (4m)(5N) + (2m)(5N)/2 W = 20J + 5J = 25 J

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Work Done By Multiple Forces the system can be modeled as a particletotal workON work done by the net force If more than one force acts on a system and the system can be modeled as a particle, the total work done ON the system is the work done by the net force(7.8)

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Work Done by Multiple Forces, cont. cannot be modeled as a particletotal work the algebraic sum If the system cannot be modeled as a particle, then the total work is equal to the algebraic sum of the work done by the individual forces

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Hooke’s Law The force exerted BY the spring is The force exerted BY the spring is F s = – kx (7.9) x x = 0 x is the position of the block with respect to the equilibrium position (x = 0) k spring constantforce constant (Units: N/m) k is called the spring constant or force constant and measures the stiffness of the spring (Units: N/m) Hooke’s Law This is called Hooke’s Law

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Hooke’s Law, cont. x positive F s negative When x is positive (spring is stretched), F s is negative x0 F s 0 When x is 0 (at the equilibrium position), F s is 0 negative F s positive When x is negative (spring is compressed), F s is positive

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Hooke’s Law, final F s opposite displacement The force exerted by the spring (F s ) is always directed opposite to the displacement from equilibrium F s restoring force F s is called the restoring force oscillate –xx If the block is released it will oscillate back and forth between –x and x

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Work Done by a Spring block system Identify the block as the system work The work as the block moves from: x i = – x max to x f = 0 (7.10)

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Work Done by a Spring, cont work The work as the block moves from: x i = 0 to x f = x max (7.10) (a)

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Work Done by a Spring, final Therefore: Net Work spring force –x max x max ZERO!!!! Net Work done by the spring force as the block moves from –x max to x max is ZERO!!!! x i x f For any arbitrary displacement: x i to x f : (7.11)

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Spring with an Applied Force F app Suppose an external agent, F app, stretches the spring applied force equalopposite spring force: The applied force is equal and opposite to the spring force: F app = – F s = – ( – kx) F app = kx

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Spring with an Applied Force, final F app Work done by F app x i = 0 to x f = x max when x i = 0 to x f = x max is: W Fapp = ½kx 2 max x i x f For any arbitrary displacement: x i to x f : (7.12)

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Active Figure 7.10

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Kinetic Energy Kinetic Energy is the energy of a particle due to its motion K = ½ mv 2 (7.15) K = ½ mv 2 (7.15) K K is the kinetic energy m m is the mass of the particle v v is the speed of the particle K:Joules (J) Units of K: Joules (J) 1 J = N m = (kg m/s 2 )m= kg m 2 /s 2 = kg(m/s) 2 change kinetic energy transfer energy A change in kinetic energy is one possible result of doing work to transfer energy into a system 7.5 Kinetic Energy And the Work-Kinetic Energy Theorem

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Kinetic Energy, cont Calculating the work: Knowing that: F = ma = mdv/dt =m(dv/dt)(dx/dx) Fdx = m(dv/dx)(dx/dt)dx = mvdv (7.14)

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Work-Kinetic Energy Theorem Work-Kinetic Energy Principle The Work-Kinetic Energy Principle states W = K f – K i = K (7.16) work done change in kinetic In the case in which work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. We can also define the kinetic energy K = ½ mv 2 (7.15) K = ½ mv 2 (7.15)

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Work-Kinetic Energy Theorem, cont Summary: m v 1 v 2 Summary: Net work done by a constant force in accelerating an object of mass m from v 1 to v 2 is: W net = ½mv 2 2 – ½mv 1 2 K “Net work on an object = Change in Kinetic Energy” It’s been shown for a one-dimension constant force. However, this is valid in general!!! It’s been shown for a one-dimension constant force. However, this is valid in general!!!

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Work-Kinetic Energy REMARKS!! W net W net ≡ work done by the net (total) force. W net scalar. W net is a scalar. W net positivenegative K +– W net can be positive or negative since K can be both + or – K ½mv 2 positiveMassv 2 (Question 10 Homework) K ½mv 2 is always positive. Mass and v 2 are both positive. (Question 10 Homework) UnitsJoules Units are Joules for both work & kinetic energy. The work-kinetic theorem: speed velocity The work-kinetic theorem: relates work to a change in speed of an object, not to a change in its velocity.

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Example 7.8 Question #14 (a). K i ½m v 2 ≥ 0 (a). K i ½m v 2 ≥ 0 K depends on: v 2 ≥ 0 & m > 0 v 2v If v 2v K f = ½m (2v) 2 = 4(½mv 2 ) = 4K i K f = ½m (2v) 2 = 4(½mv 2 ) = 4K i Doubling four times Then: Doubling the speed makes an object’s kinetic energy four times larger (b). If W = 0 v same (b). If W = 0 v must be the same at the final point as it was at the initial point

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Example 7.9 Work-Kinetic Energy Theorem (Example 7.7 Text Book) m = 6.0kg F = 12N m = 6.0kg first at rest is pulled to the right with a force F = 12N (frictionless). m Find v after m moves 3.0m Solution: Solution: normalgravitational do no work The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement W = F x = (12)(3)J = 36J W = F x = (12)(3)J = 36J W = K = ½ mv f 2 – 0 W = K = ½ mv f 2 – 0 36J = ½(6.0kg)v f 2 = (3kg)v f 2 36J = ½(6.0kg)v f 2 = (3kg)v f 2 V f =(36J/3kg) ½ = 3.5m/s V f =(36J/3kg) ½ = 3.5m/s

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Example 7.10 Work to Stop a Car W net = Fdcos180°= – Fd = – Fd W net = Fdcos180°= – Fd = – Fd W net = K = ½mv 2 2 – ½mv 1 2 = – Fd -Fd = 0 – ½m v 1 2 d v 1 2 speed doubled 4 times greater d = 80 m If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

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F –F m A moving hammer strikes a nail and comes to rest. The hammer exerts a force F on the nail, the nail exerts a force –F on the hammer (Newton's 3 rd Law) m Work done on the nail is positive: Work done on the nail is positive: K n = Fd = ½m n v n 2 – 0 > 0 W n = K n = Fd = ½m n v n 2 – 0 > 0 Work done on the hammer is negative: Work done on the hammer is negative: W h = K h = – Fd = 0 – ½m h v h 2 < 0 Example 7.11 Moving Hammer can do Work on Nail

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Example 7.12 Work on a Car to Increase Kinetic Energy W net Find W net to accelerate the 1000 kg car. W net = K = K 2 – K 1 = ½m v 2 2 – ½m v 1 2 W net =½(10 3 kg)(30m/s) 2 – ½(10 3 kg)(20m/s) 2 W net = ½(10 3 kg)(30m/s) 2 – ½(10 3 kg)(20m/s) 2 W net = 450,000J – 200,000J = 2.50x10 5 J

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Example 7.13 Work and Kinetic Energy on a Baseball 145-g 25m/sv 1 = 0 A 145-g baseball is thrown so that acquires a speed of 25m/s. ( Remember: v 1 = 0) K Find: (a). Its K. W net (b). W net on the ball by the pitcher. K ½mv 2 (a). K ½mv 2 = ½(0.145kg)(25m/s) 2 K 45.0 J W net K = – = 45.0J – 0J (b). W net = K = K 2 – K 1 = 45.0J – 0J W net = 45.0 J

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nonisolated system A nonisolated system is one that interacts with or is influenced by its environment isolated system An isolated system would not interact with its environment Work-Kinetic Energy nonisolated systems The Work-Kinetic Energy Theorem can be applied to nonisolated systems 7.6 Non-isolated System

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Internal Energy internal energy, E int The energy associated with an object’s temperature is called its internal energy, E int the surfacethe system In this example, the surface is the system The friction increasesinternal energy The friction does work and increases the internal energy of the surface

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Active Figure 7.16

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Potential Energy Potential energy configuration interact by forces Potential energy is energy related to the configuration of a system in which the components of the system interact by forces Examples include: Examples include: elastic potential energy – stored in a spring elastic potential energy – stored in a spring gravitational potential energy gravitational potential energy electrical potential energy electrical potential energy

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Ways to Transfer Energy Into or Out of A System Work Work – transfers by applying a force and causing a displacement of the point of application of the force Mechanical Waves Mechanical Waves – allow a disturbance to propagate through a medium Heat Heat – is driven by a temperature difference between two regions in space

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More Ways to Transfer Energy Into or Out of A System Matter Transfer Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it Electrical Transmission Electrical Transmission – transfer is by electric current Electromagnetic Radiation Electromagnetic Radiation – energy is transferred by electromagnetic waves

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Examples of Ways to Transfer Energy a) Work a) Work b) Mechanical Waves b) Mechanical Waves c) Heat c) Heat d) Matter transfer d) Matter transfer e) Electrical Transmission e) Electrical Transmission f) Electromagnetic radiation f) Electromagnetic radiation

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Conservation of Energy Energy is conserved Energy is conserved cannot be created or destroyed This means that energy cannot be created or destroyed If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer Mathematically: system (7.17) Mathematically: system (7.17) E system E system is the total energy of the system T T is the energy transferred across the system boundary T work = WT heat = Q Established symbols: T work = W and T heat = Q Work-Kinetic Energy theorem Conservation of Energy The Work-Kinetic Energy theorem is a special case of Conservation of Energy

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Examples to Read!!! Examples to Read!!! Example 7.7 Example 7.7 (page 195) Example 7.9 Example 7.9(page 201) Example 7.12 Example 7.12 (page 204) Homework to be solved in Class!!! Homework to be solved in Class!!! Problems: 11, 26 Problems: 11, 26 Material for the Final

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