Chapter 5

1. Chapter 5 Work and Kinetic Energy

2. Work (W) by Constant Force: • A constant force (F) acts on a body to cause displacement of the body. • The direction of motion is the same as the direction where the force applies. • W = F • s (a scalar quantity) • Where W is the work [Joule “J” = 1 N  1 m = kg m2 / s2 ] • F is the constant force [N] • s is the displacement [m]

3. F  F cos  Straight line displacement Scalar s Dot product If s = 0, W = 0 for any F A force does no work if the object does not move If  = 90, W = 0 for any F Work done by a force which is applied perpendicular to the displacement is zero Only the component of F along the displacement is doing work.

4. F F W = - Fy W = Fy  y  y Fg = mg Fg = mg • Work done is positive: • the force and the displacement in the same direction, e.g. an object is lifted by an upward force • Work done is negative: • the force and the displacement in the opposite direction, e.g. lowering an object; gravitational force

5. University Physics by Young & Freedman Zero work Positive work Positive work Negative work The net force = 0 The net force causes the speed to increase The net force causes the speed to increase The net force causes the speed to decrease (net force opposites the displacement)

6. N T f mg Example: A box is pulled up a rough inclined surface by a weight as shown in the figure. Find the numbers of forces that are doing work on the box? Ndoes no work (perpendicular to moving direction) Tdoes positive work 3 forces do work f does negative work mg does negative work Any force not perpendicular to the motion will do work

7. a f a N mg Example: A box rests on in inclined block which has a constant acceleration “a”. Find how many forces are doing work on the box? W = Fs Only forces that have a component along the direction of the displacement are doing work. 2 forces are doing work - arrows in red

8. F F s Example: A force F= 20 N pushes a wooden box across a frictionless floor for a distance s= 10 m. Find the work done by F. Work done byF on the wooden box : W = Fs = Fs (F is parallel to s, cos 0o =1) W = (20 N) x (10 m) = 200 Joules (J)

9. University Physics by Young & Freedman Work Done by a Varying Force F = kx Where F is the force pulling the string [N] k is the spring constant [N/m] x is the extension of the spring [m]

10. y spring x x1 x2 Work Done by a Spring • Hooke’s law: F = -kx (the force is a restoring force) • Work done by the spring when the block is displaced from x1to x2:

11. Energy is present in different forms and can transform from one to another: • Mechanical energy [Joules] • Electrical energy [kWh] • Thermal energy [Btu, cal] • etc • Energy • has ability to do work • is a scalar quantity • is always positive Kinetic energy: The energy associated with the speed of a moving particle Potential energy: The energy associated with the gravitational force

12. Kinetic Energy (K) Notes: 1) kinetic energy is the energy associated with the motion of an object 2) kinetic energy is not a negative value 2) kinetic energy = 0 when the particle is at rest (1) Derivation of Work-Energy Theorem using a constant net force The net force acting on an object is F. The work done over a distance s is: W = Fs = (ma)s The initial velocity of the object is v1.After moving through the distance, s, the object has the speed, v2. We have the following relationship: v22 = v12 + 2as

13. From Newton’s 2nd Law, F = ma, we obtain Work-Energy Theorem

14. When a system is accelerated from rest (V0 = 0): • Wnet = (KE)final – (KE)initial = (KE)final – 0 = (KE)final • Kinetic energy can play the roles as: • the total work to accelerate the system from rest to its final speed. • the total work of the system which can do in the process of being brought to rest.

15. Work-Energy Theorem (2) Derivation of Work-Energy Theorem using a variable net force: The work done by the net force on a particle = the change in the particle’s kinetic energy Wtot = +ve [v2 > v1, the kinetic energy increases] Wtot = -ve [v2 < v1, the kinetic energy decreases] Wtot = 0 [v2 = v1, the kinetic energy remains unchanged].

16. Point 1 Point 2 v1 v2 f f s Kinetic Friction The kinetic energy for an object with an initial velocity v1 sliding on a rough horizontal surface over a distance s before reaching a final velocity v2: The loss in kinetic energy is expressed as –fs. This is the energy dissipated by the force of kinetic friction (f) over a distance (s). Part of the energy is transferred to the internal energy of the object, and part to the surface. V1 > V2

17. N P g fK 300 mg s Example: A 1.00  102 kg box is being pulled across a horizontal floor by a force P that makes an angle of 30.0˚ above the horizontal. The coefficient of kinetic friction is 0.2. What should be the magnitude of P, so that (1) the net work done by it and (2) the kinetic frictional force is zero? x-direction: Wp = P cos 30  s (work done due to P in x-dir.) Wf = fkcos 180˚  s or (- fk  s) (work done due to f in x-dir.) y-direction: N + P sin 30o = mg fk = k N =k (mg – P sin 30o) (1) If net work done is 0, Wp+ Wf = 0 or Wp = – Wf P = 203N (2) For fk = 0, velocity is constant. No change in velocity.

18. Exercise 6-14 “University Physics” A sled with a mass of 8 kg moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4 m/s: after it has travelled 2.5m beyond this point, its speed is 6 m/s. Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled’s motion. Solution: W = Fs, K=mv2/2 and Wtot=K2 – K1 = K, we find F:

19. Power • Power is the time rate of doing work Instantaneous rate at which force F does work on an object • Unit of Power = Watt = Joule/second • 1 kWh = (103 W)(3600 s) = 3.60 x 106 J

20. Fx (N) 10 5 0 4 8 12 Exercise 6-24 “University Physics”: A child applies a force F parallel to the x-axis to a 10 kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in the figure. Calculate the work done by the force F when the sled moves a) from x = 0 to x = 8 m; b) from x = 8 m to x = 12m; c) x = 0 to 12 m. Neglect friction between the sled and the surface of the pond. Solution: The work can be found by finding the area under the graph, being careful of the sign of the force. The area under each triangle is ½ base x height. a)               ½ (8 m)(10 N) = +40 J. b)               ½ (4 m)(10 N) = +20 J. c)               ½ (12 m)(10 N) = +60 J. x (m)

21. The normal force does no work. The work-energy theorem, along with Kinetic energy equation K = mv2 / 2, gives where h = L sin  is the vertical distance the block has dropped, and  is the angle The plane makes with the horizontal. Using the given numbers, Exercise 6-20 “University Physics”: A block of ice with mass 2.0 kg slides 0.75 m down an inclined plane that slopes downward at an angle of 36.9o below the horizontal. If the block of ice starts from rest, what is the final speed? Neglect the friction. Solution:

22. Using P = F • v. Gravity is doing negative work, so the rope must do positive work to lift the skiers. The force F is gravity, and F = Nmg, where N is the number of skiers on the rope. The power is then P = (Nmg)(v) cos  = (50)(70 kg) (9.80 m/s2) (12.0 km/h) = 2.96 x 104 W. Note: the force is vertical, but the angle 15.0o is measured from the horizontal, so  = 90.0o – 15.0o is used. Exercise 6-42 “University Physics”: A ski tow is operated on a 15o slope of length 300m. The rope moves at 12 km/h and power is provided for 50 riders at one time, with an average mass per rider of 70 kg. Estimate the power required to operate the tow. Solution: