DHS 05 Work, Energy & Power (Lecture Notes & Tutorial)

Dunman High School (Senior High Physics) Topic 5-1 | Topic 5 Work, Energy, Power Guiding Questions: • What is energy and how does it relate to work done? • How does energy get transferred and transformed? Content • Work • Energy conversion and conservation • Efficiency • Potential energy and kinetic energy • Power Learning Outcomes Candidates should be able to: (a) define and use work done by a force as the product of the force and displacement in the direction of the force (b) calculate the work done in a number of situations including the work done by a gas which is expanding against a constant external pressure: W = pΔV (c) give examples of energy in different forms, its conversion and conservation, and apply the principle of energy conservation (d) show an appreciation for the implications of energy losses in practical devices and use the concept of efficiency to solve problems (e) derive, from the equations for uniformly accelerated motion in a straight line, the equation E k = 21 2 mv (f) recall and use the equation Ek = 21 2 mv (g) distinguish between gravitational potential energy, electric potential energy and elastic potential energy (h) deduce the elastic potential energy in a deformed material is related to the area under the force-extension graph (i) show an understanding of and use the relationship between force and potential energy in a uniform field to solve problems (j) derive, from the definition of work done by a force, the equation E p = mgh for potential energy changes near the Earth’s surface (k) recall and use the equation Ep = mgh for gravitational potential energy changes near the Earth's surface (l) define power as work done per unit time and derive power as the product of a force and velocity in the direction of the force.

Dunman High School (Senior High Physics) Topic 5-2 | (a) define and use work done by a force as the product of the force and displacement in the direction of the force Work is a concept that provides a link between force and energy. Do not be misled by the many different meanings of the work in ordinary conversation – doing homework, going to work or having too much work to do. Not everything we call work in conversation is work i n the physics sense of the word. We will now define work – in the physics sense. Work done (by a constant force) Work done by a constant force on an object is defined as the product of the force and its displacement in the direction of the force. W = (F cos θ)s or W = F (s cos θ) where F is the magnitude of the force, s is the magnitude of the displacement and θ is the angle between the force and the displacement. Although force and displacement are vectors, work is a scalar quantity and the SI unit of work is the joule (J). θ = 0o W = F.s θ = 180o W = θ = 90o W = W = 0 also when (i) F = 0 or (ii) s = 0 0o<θ < 90o W = 90o<θ < 180o W = Note: 1. Work is an energy transfer. If the work done on a system is positive, energy is transferred to t he system. If the work done is negative, energy is transferred from the system. 2. When discussing work, always use the phrase “Work done by ..… on …..” s F θ F, s s F s F s F θ s F θ

Dunman High School (Senior High Physics) Topic 5-3 | Example 1 Example 2 A physics book is pushed 1.20 m along a horizontal tabletop by a horizontal force of 2.00 N. The opposing force of friction is 0.400 N. (a) How much work is done on the book by the 2.00 N force? (b) What is the work done on the book by the friction force? (c) What is the total work done on the book? Solution (a) Work done by the 2.00 N force on the book = F s = (2.00)(1.20) = 2.40 J (b) Work done by the friction force on the book= − f s = (−0.400)(1. Solution A man of mass 60 kg carrying a pail of 20 kg walks on a level road for 10 m. What is the work done by the man in carrying the load? Solution Since force applied is perpendicular to the displacement, work done by the man in carrying the load = F s cos 90° = 0 J (a) Work done by W on the block = (Wsin30o)(s) = (mgsin30o)(2.50) = 61.3 J (b) Work done by f on the block = − (f.s) = − (18.5)(2.50) = − 46.3 J (c) Work done by N on the block = (N.s cos 90o) = 0 (d) Total work done on the block = (61.3 − 46.3) = 15 J

Dunman High School (Senior High Physics) Topic 5-4 | Example 3 A 5.00 kg block slides from rest down a 2.50 m long rough incline of 30° . A friction force of 18.5 N acts on the block throughout the motion. Determine (a) the work done by the force of gravity. (b) the work done by friction force between the block and the incline. (c) the work done by the normal force. (d) the total work done on the block. Note: 1 When several forces act on an object, the total (or net) work done is the sum of the work done by each force separately, that is, Wtotal = W1 + W2 + W3 + . . . = ΣWi 2 Another way to calculate the total work done is to find the work done by the net force as if there were a single force acting and then using the definition of work, that is, Wtotal = (Fnet cos θ)s where θ is the angle between Fnet and the displacement s. 30°

Dunman High School (Senior High Physics) Topic 5-5 | (b) calculate the work done in a number of situations including the work done by a gas which is expanding against a constant external pressure: W = P∆V (h) deduce the elastic potential energy in a deformed material is related to the area under the force-extension graph Graphical interpretation of work done (b) For a variable force, the work done on an object that undergoes a displacement is still equal to the area under the force−displacement graph, that is, W Fdx=∫ Example 4 The force on an object, acting along the x axis, varies as shown in the figure. Determine the work done by this force to move the object (a) from x = 0 to x = 10 m, and (b) from x = 0 to x =15 m Solution (a) Work done by F = area under force−displacement graph = )400)( 3 (2 1)400)( 4 ( )400)( 3 (2 1 ++ = 2800 J (b) Work done by F = )200)(2 (2 1)200)( 1 ( )200)( 2 (2 12800 − + − + − += 2200 J (a) If a constant force acts on an object over a displacement s, then the force-displacement graph would look like the one on the right. Work done W = Fs = area under the force-displacement graph F s work -300 -200 -100 0 100 200 300 400 500 0 5 10 15 20 F/N x/m Force F/N vs x/m F/N s

Dunman High School (Senior High Physics) Topic 5-6 | Work done to stretch a spring If the spring obeys Hooke’s law, F = ke, that is, the displacement e of the spring is proportional to the force F applied on it. The work done W by the force in stretching the spring is given by the area under the force−extension graph. This work done is stored as elastic potential energy U in the spring. U = Fe2 1 = 2 2 1 ke Work done by a gas during expansion Consider a quantity of gas, at pressure p, in a syringe which has a frictionless piston as shown in the diagram. If the piston has cross-sectional area A, the force exerted by the gas on the piston is F = pA If the gas expands slowly (so that we may consider the pressure of the gas to be constant) against a constant external pressure moving outwards a displacement s, then the force F is constant. The work done by the gas in expanding from V1 to V2 W = F.s = (pA)s = p(As) = p∆V = p(V2 − V1 ) Note:  When the gas expands (V2 > V1), work done by the gas W is positive.  When the gas is compressed (V2 < V1), work done by the gas W is negative.  Graphically, work done ∫= pdVW , i.e.,