Mechanics & Properties of Matter 5: Energy and Power Energy and Power AIM This unit re-introduces the formulae for calculating work done, potential energy, kinetic energy and power. The principle that the total energy is conserved during an energy transformation is used to help solve problems involving objects on an inclined plane, etc. OBJECTIVES On completing this unit you should be able to: carry out calculations involving work done, potential energy, kinetic energy and power. Strathaven Academy - 1 - Mechanics and Properties of Matter
Work Done (E w ) Work is done whenever an object is moved against a force or resistance. When work is done, energy is transformed. e.g. When work is done against the force of friction, kinetic energy is transformed into heat energy. Work done = force x distance or, E w = Fd The unit of work done is the newton.metre or the joule (J). The joule is also the unit of energy. Potential Energy (E p ) When an object is lifted against the pull of gravity, work has to be done and energy is transformed into gravitational potential energy. Potential energy gained = work done against gravity or, E p = mgh Kinetic Energy (E k ) When work is done to overcome the reluctance of a mass to accelerate, energy is transformed into kinetic energy. E k = ½ mv 2 Power (P) Power is the rate of doing work (or of transforming energy). Power = work done time or, P = E w /t When work is done at the rate of one joule per second, the power is one watt (W). Strathaven Academy - 2 - Mechanics and Properties of Matter
Conservation of energy Energy can never be created nor destroyed - it s only ever transformed from one form to another. The above statement is called the principle of conservation of energy. The principle means that, even when energy seems to be lost, it is never destroyed. It is simply transformed into a new form. Sometimes the new form is not as obvious as the old one, so you may not notice it! This is illustrated by the following example. When you apply the brakes on your bike, the kinetic energy is not destroyed it is transformed into heat in the brakes. In this case the temperature rise may be too small for you to notice - but the brakes on grand prix racing cars often glow red hot, showing the principle to be true! Calculating the speed of falling objects When an object falls, its potential energy is transformed into kinetic energy (by the principle of conservation of energy). So the further it falls, the faster it goes. In practice, some of the potential energy will be transformed into heat due to air resistance. In exam questions you may be told to ignore air resistance, in which case you can use this simple formula: kinetic energy gained = potential energy lost Example: A stone of mass 2.0 kg is dropped from a height of 5.0 m. Calculate its speed as it hits the ground. Solution: Step 1) Find the E p lost: E p = mgh = 2.0 x 9.8 x 5.0 = 98 J Step 2) Apply the equation: E k gained = E p lost ½ mv 2 = 98 ½ x 2.0 x v 2 = 98 v 2 = 98 Step 3) Take the square root: v = 98 v = 9.9 ms -1 Energy Transformations -Objects on an inclined plane Strathaven Academy - 3 - Mechanics and Properties of Matter
1. No friction (e.g. skier on snow slope) u = 0 m h v On the way down the slope, the skier gathers speed as his gravitational potential energy is transformed into kinetic energy. The principle that the total energy is conserved during an energy transformation leads to the following equation: loss of potential energy = gain of kinetic energy mgh = ½ m v 2 N.B. Since m is the same on both sides of the equation: gh = ½ v 2 This allows v (the final speed) to be calculated knowing the height of the slope. Note, if there is no friction, that the mass of the skier does not affect his final speed, only the height of the slope h. 2. With friction (e.g. packing case on ramp) d h When friction is present, some energy is converted into heat energy. This means that the kinetic energy at the bottom, and hence the final speed v, will be less than in situation 1 where there is no friction. loss of E p = gain of E k + work done against friction mgh = ½ m v 2 + Fd where F = the force of friction Example A trolley is released down a slope from a height of 0.3 m. If its speed at the bottom is found to be 2 ms -1, find a) the energy difference between the E p at top and E k at the bottom. Strathaven Academy - 4 - Mechanics and Properties of Matter
b) the work done by friction c) the force of friction on the trolley 2 m 1 k g h = 0. 3 m a) E p at top = mgh = 1 9.8 0.3 = 2.94 J E k at bottom = ½mv 2 = ½ 1 4 = 2 J Energy difference = 0.94 J b) Work done by friction = energy difference (due to heat, sound) = 0.94 J c) Work done = Force of friction d = 0.94 J d = 2 m F = 0.94 = 0.47 N 2 Force of friction = 0.47 N Power Power is the rate of transformation of energy from one form to another. P = energy time = work done time = F displacement t = F average velocity Power is measured in watts (W). Strathaven Academy - 5 - Mechanics and Properties of Matter
Work done, energy and power 1. A train is travelling at a constant speed of 30 ms -1 when the driver spots some obstruction on the track ahead. He immediately pulls the emergency brakes and the train comes to rest in 8 seconds. The mass of the train is 80 000 kg and the average braking force is 750 000 N. (a) How much kinetic energy did the train have before braking? (b) What happened to this kinetic energy during braking? (c) The obstruction, a fallen tree, was 125 m from the train at the instant the driver pulled the brakes. Did the train collide with the tree? Justify your answer. (d) Calculate the braking power of the train. 2. A box of mass 8 kg is dragged 7m along a floor in a warehouse against a constant frictional force of 20 N. It is then lifted onto a shelf which is 1.2 m above the floor. (a) Calculate the total energy required for this operation. (b) If it takes 1 minute to complete the job, what is the average power output of the warehouse worker? 3. A soldier of mass 70 kg climbs a rope on an assault course. It takes him 50 seconds to climb the rope and he has a power output of 160 W. Calculate the height to which he climbed. Strathaven Academy - 6 - Mechanics and Properties of Matter
4. A cyclist on a cycle path stops at the top of a slope and decides to free wheel down it. The slope is 8 m high and 14 m long, and the combined mass of cyclist plus bicycle is 80 kg. An average force of 300 N opposes the cyclist as he travels down the slope. This is due to friction and air resistance. (a) How much gravitational potential energy does the cyclist lose as he travels down the slope? (b) How much work does the cyclist do against friction and air resistance on the slope? (c) How much kinetic energy does the cyclist have at the foot of the slope? (d) Calculate the speed of the cyclist at the foot of the slope. (e) If friction and air resistance amount to 320 N on the flat surface how far will the cyclist travel before these frictional forces bring him to rest? 5. A coach of mass 5 000 kg is travelling along a motorway at 70 m.p.h. (31 3 ms -1 ) when the driver notices roadworks ahead. He brakes to a speed of 45 m.p.h. (20 1 ms -1 ). The graph below shows how the speed of the coach changed as it approached the roadworks. speed in ms -1 31 3 20 1 0 1 2 3 4 5 time in seconds (a) Calculate the loss of kinetic energy of the coach during braking. (b) Calculate the braking power of the coach. Strathaven Academy - 7 - Mechanics and Properties of Matter