Work, Energy and Power

Work = Force component x displacement Work = F x x 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 30 o 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 30 o with the horizontal? Work =(Fcosθ)x = (60 N)(cos30 o )(50m) Work = 2600 Nm

Resultant Work When work involves several forces we can calculate the resultant work 2 ways 1 st – Find the work done by each force and add them 2 nd – Find the resultant force and multiply by the displacement

Example 8.2 A push of 80 N moves a 5 kg block up a 30 o 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 μ k N= 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 = 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 –F f –W x =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 mv 2 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 mv 2 = 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 = 1/2mv f 2 – 1/2mv i 2

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(v f ) 2 or Fx = -1/2mv f 2 F = -mv f 2 /2x = 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/s 2 )(2 m) = 23.5 J The potential energy relative to the floor is U = mgh= (1.2 kg)(9.8 m/s 2 )( 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 mv 2 mgh = ½ mv2 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 30 o 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 mgh 0 + ½ mv o 2 = mgh f + ½ mv f 2 +F f x Using v o = 0, and Ff = µkN= (0.2)(170 N) We get 7840 J = ½ (20 Kg)v f J v f = 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 10 5 W = 122 kW