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Work, Energy and Power Work, Energy and Power • Work = Force component x displacement • Work = Fxx • When the displacement is perpendicular to the force, no work is done. • When there is no displacement, no work is done Example 8.1 • What work is done by a 60-N force pulling the wagon below a distance of 50 meters when the force transmitted by the handle makes an angle of 30o with the horizontal? Example 8.1 • What work is done by a 60-N force pulling the wagon below a distance of 50 meters when the force transmitted by the handle makes an angle of 30o with the horizontal? • Work =(Fcosθ)x = (60 N)(cos30o)(50m) • Work = 2600 Nm Resultant Work • When work involves several forces we can calculate the resultant work 2 ways • 1st – Find the work done by each force and add them • 2nd – Find the resultant force and multiply by the displacement Example 8.2 • A push of 80 N moves a 5 kg block up a 30o inclined plane, as shown below. The coefficient of kinetic friction is 0.25, and the length of the plane is 20 m. • (a) Compute the work done by each force acting on the block • (b)Show that the net work done by these forces is the same as the work done by the resultant force Example 8.2 • Consider the work done by each force • Work done by the normal force n is 0 • Work done by the pushing force P is (80 N)(1)(20 m) = 1600 J • The friction force is μkN = -10.6 N, the negative sign indicates this force opposes the motion. The work done by the friction force is (-10.6 N)(20 m) = -212 J • Work done by x component of the weight = -(24.5 N)(20 m) = -490 N • Net work = 0 + 1600 J – 212 J -490 J = 898 J Example 8.2 • Determine the resultant force and calculate the work done by the resultant force • Fr = P –Ff –Wx • =80N – 10.6N -24.5N = 44.9N • Net work is • (44.9N)(20 m) = 898 J Energy • Energy may be thought of as anything that can be converted to work. • We will consider two kinds of energy • Kinetic Energy – Energy possessed by a body by virtue of its motion • KE = ½ mv2 • Potential Energy – Energy possessed by a system by virtue of its position or condition. • Gravitational Potential Energy, U = mgh Kinetic Energy • Consider the following examples of objects with kinetic energy: • • • • 1200 kg car, v = 80 km/h, (22.2 m/s) KE = ½ (1200 kg)(22.2 m/s)2 = 296,000 J 20 g bullet v = 400 m/s KE = ½ (0.02 kg)(400 m/s)2 = 1,600 J Example 8.3 • Find the kinetic energy of a 4 kg sledgehammer with a velocity of 24 m/s. • K = ½ mv2 = ½(4 kg)(24 m/s)2 • K = 1150 J Work-Energy Theorem • When work is done on a mass to change its motion, The work done is equal to the change in kinetic energy • Fx = 1/2mvf2 – 1/2mvi2 Example 8.5 • What average force is necessary to stop a 16 gram bullet traveling at 260 m/s as it penetrates 12 cm into a block of wood? • Fx = 1/2m(0)2 – ½ m(vf)2 or Fx = -1/2mvf2 • F = -mvf2/2x = -4510 N Potential Energy • The energy that systems possess by virtue of their positions or conditions is potential energy. • When work is done against the force of gravity, it gives the object gravitational potential energy • Gravitational potential energy is found using U = mgh Potential Energy • Lifting a mass to a height h requires work mgh. If the mass falls it acquires kinetic energy equal to its potential energy Example 8.6 • A 1.2 kg toolbox is held 2 meters above the top of a table that is 80 cm from the floor. Find the potential energy relative to the top of the table and relative to the floor. • The potential energy relative to the tabletop is: • U =mgh =(1.2 kg)(9.8 m/s2)(2 m) = 23.5 J • The potential energy relative to the floor is • U = mgh= (1.2 kg)(9.8 m/s2)(2.0 +0.8 m) = 32.9 J Conservation of Energy • In the absence of air resistance or other dissipative forces, the sum of the potential and kinetic energies is a constant. Provided that no energy is added to the system. Conservation of Energy As the ball the falls from height h, the sum of potential, U and Kinetic, K energies is conserved. Example 8.8 • In the figure below, a 40 kg wrecking ball is pulled to one side until it is 1.6 m above its lowest point. Neglecting friction, what will be its velocity as it passes through its lowest point? Example 8.8 • Applying Conservation of Energy • mgh + 0 = 0 + ½ mv2 • mgh = ½ mv2 v 2 gh 2(9.8m/s )(1.6m 2 • V = 5.6 m/s Energy and Friction Forces • When dissipative forces are present, they must be included in the final energy total • Initial total energy = final total energy + losses due to friction Example 8.9 • A 20 kg sled rests at the top of a 30o slope 80 meters in length, as shown below. • If µk = 0.2, what is the velocity at the bottom of the incline? Example 8.9 • Using the Conservation of Energy equation, we get • mgh0 + ½ mvo2 = mghf + ½ mvf2 +Ffx • Using vo = 0, and Ff = µkN= (0.2)(170 N) • We get 7840 J = ½ (20 Kg)vf2 + 2720 J • vf = 22.6 m/s Power • Power is the rate at which work is accomplished. • P = work/time • The unit of power is the watt, 1kilowatt = 1000 watts, 1 hp =746 watts Example 8.10 • A loaded elevator has a total mass of 2800 kg and is lifted to a height of 200 m in a time of 45 seconds. Calculate the average power. • P = Fx/t = mgh/t • P = (2800 kg)(9.8 m/s2)(200 m)/45 s • P = 1.22 x 105 W = 122 kW