7.4 Work Done by a Varying Force

1. 7.4 Work Done by a Varying Force

2. Work Done by a Varying Force • Assume that during a very small displacement, Dx, F is constant • For that displacement, W ~ FDx • For all of the intervals,

3. 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!!

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

5. Work Done By Multiple Forces • 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)

6. Work Done by Multiple Forces, cont. • 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

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

8. Hooke’s Law, cont. • When xis positive (spring is stretched), Fsisnegative • When x is 0 (at the equilibrium position), Fs is 0 • When x is negative(spring is compressed), Fsis positive

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

10. Work Done by a Spring • Identify the block as the system • The work as the block moves from: xi = –xmaxtoxf = 0 (7.10)

11. Work Done by a Spring, cont • The work as the block moves from: xi = 0 toxf = xmax (7.10) (a)

12. Work Done by a Spring, final • Therefore: Net Workdone by the spring force as the block moves from –xmax toxmaxisZERO!!!! • For any arbitrary displacement: xito xf: (7.11)

13. Spring with an Applied Force • Suppose an external agent, Fapp, stretches the spring • The applied force is equal and opposite to the spring force: Fapp=–Fs=–(–kx)Fapp = kx

14. Spring with an Applied Force, final • Work done by Fapp when xi = 0 toxf = xmaxis: WFapp = ½kx2max • For any arbitrary displacement: xito xf: (7.12)

15. Active Figure 7.10

16. 7.5 Kinetic Energy And the Work-Kinetic Energy Theorem • Kinetic Energy is the energy of a particle due to its motion • K = ½ mv2 (7.15) • K is the kinetic energy • m is the mass of the particle • v is the speed of the particle • Units of K:Joules (J) 1 J = N•m = (kg•m/s2)m= kg•m2/s2 = kg(m/s)2 • A change in kinetic energy is one possible result of doing work to transfer energy into a system

17. 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)

18. Work-Kinetic Energy Theorem • The Work-Kinetic Energy Principle states SW = Kf– Ki= DK (7.16) • 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 mv2 (7.15)

19. Work-Kinetic Energy Theorem, cont • Summary: Net work done by a constant force in accelerating an object of mass m from v1to v2is: Wnet = ½mv22 – ½mv12 DK “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!!!

20. Work-Kinetic Energy REMARKS!! • Wnet≡work done by the net (total) force. • Wnetis a scalar. • Wnetcan be positive or negative since Kcan be both + or – • K  ½mv2is always positive. Mass and v2are both positive. (Question 10 Homework) • Unitsare Joules for both work & kinetic energy. • The work-kinetic theorem:relates work to a change in speedof an object, not to a change in its velocity.

21. Example 7.8Question #14 • (a). Ki ½m v2 ≥ 0 Kdepends on:v2 ≥ 0 &m > 0 Ifv2v Kf = ½m (2v)2 = 4(½mv2 ) = 4Ki • Then: Doubling the speed makes an object’s kinetic energyfour timeslarger • (b). If SW = 0  vmust be the same at the final point as it was at the initial point

22. Example 7.9Work-Kinetic Energy Theorem (Example 7.7 Text Book) • m = 6.0kg first at rest is pulled to the right with a force F = 12N (frictionless). • Find v after m moves 3.0m • Solution: • The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement • W = F Dx =(12)(3)J = 36J • W = DK = ½ mvf2– 0 • 36J = ½(6.0kg)vf2 = (3kg)vf2 • Vf =(36J/3kg)½ =3.5m/s

23. Example 7.10Work to Stop a Car • Wnet = Fdcos180°= –Fd= –Fd Wnet = K= ½mv22 – ½mv12 = –Fd  -Fd = 0 – ½m v12  dv12 • If the car’s initial speed doubled, the stopping distance is 4 times greater. Then: d = 80 m

24. Example 7.11Moving Hammer can do Work on Nail • A moving hammer strikes a nail and comes to rest. The hammer exerts a force Fon the nail, the nail exerts a force –F on the hammer (Newton's 3rd Law) m • Work done on the nail is positive: Wn= Kn = Fd = ½mnvn2 – 0 > 0 • Work done on the hammer is negative: Wh= Kh = –Fd = 0 – ½mhvh2 < 0

25. Example 7.12Work on a Car to Increase Kinetic Energy • Find Wnetto accelerate the 1000 kg car. Wnet= K = K2 – K1 = ½m v22 – ½m v12 Wnet=½(103kg)(30m/s)2 – ½(103kg)(20m/s)2  Wnet=450,000J – 200,000J = 2.50x105J

26. Example 7.13Work and Kinetic Energy on a Baseball • A 145-g baseball is thrown so that acquires a speed of 25m/s. ( Remember: v1 = 0) Find: (a). Its K. (b). Wneton the ball by the pitcher. • (a). K  ½mv2= ½(0.145kg)(25m/s)2 K  45.0 J • (b). Wnet = K = K2 – K1 = 45.0J – 0J  Wnet= 45.0 J

27. 7.6 Non-isolated System • A nonisolated system is one that interacts with or is influenced by its environment • An isolated system would not interact with its environment • The Work-Kinetic Energy Theorem can be applied to nonisolated systems

28. Internal Energy • The energy associated with an object’s temperature is called its internal energy, Eint • In this example, the surface is the system • The friction does work and increases the internal energy of the surface

29. Active Figure 7.16

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

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

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

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

34. Conservation of Energy • Energy is conserved • 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: SEsystem = ST (7.17) • Esystem is the total energy of the system • T is the energy transferred across the system boundary • Established symbols: Twork = W and Theat = Q • The Work-Kinetic Energy theorem is a special case of Conservation of Energy

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