Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Sub-Topics • Work and standard units •Power concept and Power calculation •Kinetic energy concept • Potential energy •The law of conservation of energy Work • an activity involving a force and movement in the direction of the force or when a force acts upon an object to cause a displacement of the object • Work Done by a constant force: product of the magnitudes of the displacement and the component of the force parallel to that displacement. W = F|| d = (F cos θ )d F = force (N) d = displacement (m) W = work (Nm @ J) θ: angle between displacement and force vectors (a) If there is no displacement, no work is done: W = 0. (b) For a constant force in the same direction as the displacement, W = Fd. (c) For a constant force at an angle to the displacement, W = (F cos θ)d. EXAMPLE 1 A students holds a book, which has a mass of 1.5 kg, out a second-story dormitory until her arm is tired, then she releases it (a)How much work is done on the book by the student in simply holding it out the window? (b)How much work is done by the force of gravity during the time in which the book falls 3.0 m? Solution Given vo = 0 (initially at rest) m = 1.5 kg d= 3.0 m (a)W = 0 (She does no work on the book, even though she exerts an upward force on the book, but the displacement is zero) (b) W = F . d = (m. g)d = (1.5kg)(9.8m/s2)(3.0m) = 44 J (While the book is falling, the force acting is gravity, which is equal in magnitude to the weight of the book) Work, Force & Displacement Total or Net work • Work done by all the forces acting on the object or scalar sum of all those quantities of work EXAMPLE 2 A 0.75kg block slides with a uniform velocity down a 20o inclined plane (refer Figure below) (a)How much work is done by force of friction on the block as it slides the total length of the plane (b) What is the net work on the block? Solution Given: m = 0.75kg θ = 20o (a) W f = f k (cos180o )d = − f k d L = 1.2 m The angle 180o indicates the force and displacement are in opposite directions Distance: d= L cos θ f k = mg sin θ Friction force: So, work done by frictional force: L W f = − (mg sin θ ) = − mgL tan θ cos θ W f = − (0.75kg )(9.8m / s 2 )(1.2m)(tan 20o ) = − 3.2 J Solution (b) To find net work, we need to calculate the work done to the object by all forces L Distance: d= cos θ Gravity force: f g = mg sin θ So, work done by gravity force: L Wg = (mg sin θ ) = mgL tan θ = + 3.2 J cos θ Net Work: Wnet = Wg + W f = + 3.2 J + − 3.2 J = 0 Work Done by a Variable Force • Forces in practice will often vary, they change in magnitude or angle with time or position. E.g. Someone might push harder and harder on an object to overcome the force of static friction, until the applied force exceeds fsmax • Example of variable force is the force applied to the spring (refer fig 5) Fs = − k∆ x = − k(x − xo ) • For ideal spring force: • Thus, W= Fs = − kx 1 2 kx 2 where k = force constant x= distance Figure 5 (a) An applied force Fa stretches the spring, and the spring exerts an equal and opposite force Fs on the hand. (b) The magnitude of the force depends on the change Δx in the spring’s length. This change is measured from to the end of the unstretched spring at x0. Note: In Fig. 5, the hand applies a variable force Fa in stretching the spring. At the same time, the spring exerts an equal and opposite force Fs on the hand. EXAMPLE 3 A 0.15kg mass is attached to a vertical spring and hangs at rest a distance of 4.6cm below its original position. An additional 0.5kg mass is then suspended form the first mass and allowed to descend to a new equilibrium position. What is the total extension of the spring? Solution Given: m1 = 0.15kg m2 = 0.5kg For ideal spring force: x1 = 4.6cm = 0.046m x=? Fs1 = kx1 = m1 g Solving to find spring constant, k: m1 g 0.15kg (9.8m / s 2 ) k= = = 32 N / m x1 0.046m Thus, Fs = (m1 + m2 ) g = kx (m1 + m2) g (0.15kg + 0.5kg )(9.8m / s 2) x= = = 0 .2 m k 32 N / m Energy • the ability to do work. When work is done, energy is transferred • forms of energy e.g. electrical, chemical heat, nuclear, mechanical etc • Units: Joules (J) • Types of energy – Kinetic – Potential • Energy comes in many forms and always conserved. Forms of Energy Radiant Thermal Electrical Nuclear Chemical Sound Mechanical Mechanical Magnetic Kinetic energy concept • Kinetic energy is the ability to do work through motion. Also called as energy of motion Remember 2 2 2 2 v f = vo + 2 a ( ∆ x ) vf v − o = a∆ x 2 2 Multiply both side with mass; 1 1 2 2 mv f − mvo = ma∆ x 2 2 Thus, kinetic energy; K= 1 2 mv 2 K = kinetic energy m= mass v = velocity vf = final velocity vo = initial velocity The work done on a block by a constant force in moving it along a horizontal frictionless surface is equal to the change in the block’s kinetic energy: W = ΔK. EXAMPLE 4 What is the kinetic energy of a 4 kg shot-put thrown by an athlete at a speed of 15 m/s? Solution Kinetic energy = 1/2 mv2 = 1/2 x 4 kg x (15 m/s)2 = 1/2 x 4 kg x 225 m2/s2 = 450 J Potential Energy • Potential energy exists whenever an object which has mass has a position within a force field (gravitational, magnetic, electrical). • An object having potential energy has potential to do work • Often called the energy of position • We will focus primarily on gravitational potential energy (energy an object has because of its height above the Earth) gravitational potential energy Gravitational potential energy U = mgy m = mass g = acceleration due to gravity y =distance Kinetic & Potential Energy Relationship If force depends on distance, ∆P E = −F ∆x For gravity (near Earth’s surface) ∆P E = m gh EXAMPLE 5 What is the potential energy of a 12 kg mass raised to a height of 25 m? Solution Potential Energy = weight x height change Weight (m x g) = 12 kg x 10 N/kg = 120 N Height change, y = height at end - height at start = 25 - 0 = 25 m Potential energy (U) = m x g x y = 120 N x 25 m = 3000 J EXAMPLE 6 A diver of 75 kg drops from a board 10.0 m above the water surface, as in the Figure. Find his speed 5.00 m above the water surface. Neglect air resistance. 9.9 m/s Potential Energy The law of conservation of energy • The Law of Conservation of Energy States: the total energy of an isolated system is always conserved • Energy is neither created nor destroyed; it is converted from one form to another • example in a nuclear power station. – Nuclear energy is converted into heat. – Heat boils water to steam – Heat in the steam is converted into kinetic energy in the turbines – Which is converted into electrical energy in the generator Conservative & Nonconservative Force Conservative Nonconservative Work done by it in moving an object is independent of the object’s path Work done by it in moving an object does depend on the object’s path depends only on the initial and final positions of an object E.g: Friction, a longer path would produce more work done by friction than a shorter one, more energy would be lost Total energy is constant E = Eo, K + U = Ko - Uo Total energy is not constant E = Eo Allows to conserve or store energy as potential energy EXAMPLE 8 A skier with a mass of 80kg starts from rest at the top of a slope and skis down from an elevation of 110m. The speed of the skier at the bottom of the slope is 20m/s (a)Show that the system is nonconservative (b)How much work is done by the nonconservative force of friction Solution Given; m = 80kg vo = 0 v = 20m/s yo = 110m (a) If the system is conservative, total mechanical energy is constant Energy at the top Eo = U = mgyo = (80kg)(9.8m/s2)(110m) = 8.6 x104 J Energy at the bottom 1 1 E = K = mv 2 = (80kg )(20m / s ) 2 = 1.6 x10 4 J 2 2 E ≠ Eo , so the system is nonconservative. (b) The amount of work done by the nonconservative force of friction is equal to the change in the mechanical energy, or amount of mechanical energy lost Wnc = E − Eo = (1.6 x10 4 J ) − (8.6 x10 4 J ) = − 7.0 x10 4 J Power Concept • • • • the rate of doing work or the rate of using energy Power = energy transferred (J) = work done (J) time taken (s) time taken (s) SI Unit : 1 W (watt) = 1 Js-1 (joule per second) British Unit: 1 hp (Horsepower) = 746 W • Work done = force x distance moved: W = Fs • Power = energy ÷ time: P = W/t • Speed = distance ÷ time: v = s/t • So we can write: P = W/t = Fs/t • Therefore: Power (W) = force (N) × speed (m/s) EXAMPLE 9 A crane hoist lifts a load of 1000kg a vertical distance of 25m in 9 s at a constant velocity. How much useful work is done by the hoist each second? Solution Given: m = 1000kg d = 25m P= t = 9.0 s W Fd mgd = = t t t (1000kg )(9.8m / s 2 )(25m) = = 27 x10 4 W = 27 kW 9.0 s EXAMPLE 10 The motors of two vacuum have net power outputs of 1.00 hp and 0.50 hp, respectively. (a)How much work in joules can each motor do in 3.0 min? (b)How long does each motor take to do 97.0 kJ of work Solution Given; P1= 1.00 hp = 746 W P2 = 0.50 hp = 373 W t = 3 min = 180 s W = 97.0 kJ (a) Since P = W/t W1 = P1t = (746 W)(180s) = 1.34 x 105 J and W2 = P2t = (373W)(180s) = 0.67 x 105 J (b) Times, t = W/P t1 = W/P1 = (97 x 103 J)/(746W) = 130 s and t2 = W/P2 = (97x 103 J)/(373 W) = 260 s Note that, smaller motor takes more time and does less work than larger motor Efficiency Relates work output to energy (work) input as a percent ε = Work output ( x100% ) = Wout ( x100% ) Energy input Ein ε = Power output ( x100% ) = Pout ( x100% ) Pin Power input EXAMPLE 11 The motor of an electric drill with an efficiency of 80% has a power input of 600kW. How much useful work is done by the drill in 30s? Solution Given; ε = 80 % Pin = 600kW t = 30s Pout = εPin = (0.8)(600kW) = 480 kW From, P = W/t wout = Poutt = (480 kW)(30 s) = 14.4 x 106 J