1 Work When a force moves something, work is done. Whenever work is done, energy is changed into a different form. Chemical energy → Kinetic energy

2 Work done by a constant force Work = force in the direction x displacement of displacement W = Fs s FF

3 force in the direction of displacement Where the force and displacement are not in the same direction, the component of force in the direction of displacement is used. s FF F cos  F sin   W = (F cos  ) s = Fs cos 

4 Positive, zero and negative work done A block is initially at rest and placed on a smooth floor. It is pushed by a horizontal force of 5 N for 2 m. Work done by the force = 5 x 2 = 10 J. Work done = + ve ⇒ mechanical energy of the block is increased due to the force 2 m 5 N

5 A 5-kg suitcase is carried by a man on his shoulder for 3 m. Work done by the force = (50 cos 90 o )(3) = 0 J Work done = 0 ⇒ mechanical energy of the block remains unchanged due to the force 50 N 3 m

6 A car is traveling and brakes are applied to stop the car. The braking force is 7500 N and the braking distance is 40 m. Work done by the force = x 40 = J - ve Work done = - ve decreased ⇒ mechanical energy of the block is decreased due to the force 7500 N 40 m

7 Work done by a varying force  W is the work done by F during a very small displacement  x. ⇒  W = F  x The total work done = sum of all work done during all small displacements = ∑F i  x i = ∫Fdx = Area under force – displacement graph. Force Displacement F  x s0 WW

8 Energy Energy exists in many different forms. But we shall only study the different forms of mechanical energy. Energy Elastic potential energy Gravitational potential energy kinetic energy

9 Kinetic Energy A moving object has kinetic energy. Consider a body of mass m which is initially at rest. Let a constant force F act on it over a distance s and bring it to move with velocity v. Since the initial velocity = 0, by equation of motion, we have 2as = v 2 – 0 2. Therefore, a = v 2 /2s. Kinetic energy of the body = Work done by F = Fs = (ma)s = m(v 2 /2s)s = ½ mv 2. F at rest v F s mm

10 Kinetic Energy from u to v In general, if the velocity of a body of mass m increases from u to v when work is done on it by a constant force F acting over a distance s, initial velocity = u v 2 – u 2 Since the initial velocity = u, by equation of motion, we have 2as = v 2 – u 2. Therefore, a = (v 2 – u 2 )/2s. Work done Kinetic energy gained by the body = Work done by F v 2 – u 2 – ½ mu 2 = Fs = (ma)s = m[(v 2 – u 2 )/2s]s = ½ mv 2 – ½ mu 2. Fuv F s mm

11 Gravitational potential energy Gravitational potential energy is the energy an object possesses because of its position above the ground. Consider an object of mass m being lifted vertically for a height h from the ground. h F F mg mg If the potential energy at the ground surface is taken to be zero, potential energy at the height h above the ground work done = work done by the force = Fs = mgh

12 Elastic potential energy Hooke’s Law: FxFx For an elastic string or spring, the extension x is directly proportional to the applied force F if the elastic limit is not exceeded. i.e. F ∝ x or F = kx where k is the force constant Extension x Applied force F Natural length l If k = 100 N m -1, find the tension if the extension is 5 cm. T = k x = (100)(0.05) = 5 N

13 Elastic potential energy Elastic potential energy in the string work done = work done by the force to achieve an extension x = Area under the F – t graph = ½ Fx = ½ (kx)x = ½ kx 2 Extension x Applied force F Natural length l Force Extension F x F ∝ x F = kx

14 Conservation of energy Energy may be transformed from one form to another, but it cannot be created or destroyed, i.e. the total energy of a system is constant. The total amount of mechanical energy (K.E. + P.E.) is constant unless the motion is frictionless. i.e. K.E. lost = P.E. gained or P.E. lost = K.E. gained smooth Mechanical energy is conserved P.E. lost = K.E. gained rough Mechanical energy not conserved P.E. lost > K.E. gained

15 A block of mass 5 kg slides down an incline plane from rest. If the angle of inclination is 30 o and the coefficient of kinetic friction  between the block and the plane is 0.2. (a)Determine work done by the gravitational force if the distance traveled is by the block is 3 m. (b)Determine the corresponding work done by the friction. (c)Hence, find the speed attained by the block. 30 o

16 Solution: (a)work done by the gravitational force = mg sin 30 o x 3 = 75 J (P.E. loss by the block) (b)frictional force =  R = 0.2 x mg cos 30 o = 8.66 N work done by friction = x 3 = J byagainst (Note: work done by friction = J; work done against friction = 26.0 J) (c)K.E. gained = P.E. loss – work done against friction = 75 – 26.0 = 49 J ½ mu 2 ½ mv 2 – ½ mu 2 = 49 ½ (5)v 2 – 0 = 49 v = 4.43 ms -1 mg cos 30 o mg sin 30 o mg R f 30 o

17 One end of an elastic string is connected to a fixed point A and the other end is connected to an object of mass 2 kg as shown. If the object is released from A, find the extension of the string when the object is instantaneously at rest. It is given that the natural length of the string is 30 cm and the force constant is 100 Nm -1. AA t = 0 Natural length = 30 cm Extension x ? Instantaneously at rest

18 Solution: By conservation of energy, P.E. lost = Elastic P.E. gained 2(10)(0.3 + x) = ½(100)x 2 50x 2 – 20x – 6 = 0 x = 0.6 m or x = -0.2 m (rejected)AA t = 0 Natural length = 30 cm Extension x ? Instantaneously at rest

19 Conservative force Conservative force is a force whose work is determined only by the final displacement of the object acted upon. The total work done by a conservative force is independent of the path taken. i.e. W Path 1 = W Path 2 a bPath 1 Path 2

20 Conservative force For example, if a child slides down a frictionless slide, the work done by the gravitational force on the child from the top of the slide to the bottom will be the same no matter what the shape of the slide; it can be straight or it can be a spiral. The amount of work done only depends on the vertical displacement of the child. Work done by gravitational force = mg sin  s = mgh (independent of the angle of the slide) h h  s mg

21 Conservative forceNon-conservative force Gravitational forceFrictional force Electrostatic force Air resistance

22 Power The rate at which work is done or energy is transferred. s FF Average power = work done / time taken P = (Fs)/t = F(s / t) = Fv If the force acts on the body at an angle  in direction of the motion, s FF F cos  F sin   P = (Fs cos  )/t = Fcos  (s / t) = F cos  v

23 Each time the heart pumps, it accelerates about 20 g of blood from 0.2 ms -1 to 0.34 ms -1. Each time the heart pumps, it accelerates about 20 g of blood from 0.2 ms -1 to 0.34 ms -1. (a)What is the increase in kinetic energy of the blood with each beat? (b)Calculate the power of the heart when it beats at about 70 times per minute. Solution: (a)Increase in K.E. = ½ mv 2 – ½ mu 2 = ½ (0.02)(0.34) 2 – ½ (0.02)(0.2) 2 = 7.56 x J (b)Power of the heart = energy / time = (7.56 x x 70) / 60 = 8.82 x W

24 Efficiency