WORK In order for work to be done, three things are necessary: There must be an applied force. The force must act through a certain distance, called the displacement. The force must have a component along the displacement.

Work is a scalar quantity equal to the product of the magnitudes of the displacement and the component of the force in the direction of the displacement.   W = F . r W = F cos  r UNITS: N.m, this unit is called a Joule (J)

If the force acting on an object varies in magnitude and/or direction during the object’s displacement, graphical analysis can be used to determine the work done. F is plotted on the y-axis and the distance through which the object moves is plotted on the x-axis. The work done is represented by the area under the curve.

6.1 What work is done by a 60 N force in dragging the block a distance of 50 m when the force is transmitted by a rope making an angle of 30 with the horizontal? F = 60 N, 30˚ r = 50 m W = F cos  r = 60 cos 30˚ (50) = 2598 J

Force needed is equal to Fg W = mgh = 3(9.8)(0.4) = 12 J 6.2 How much work is done against gravity in lifting a 3.0 kg object through a vertical distance of 40 cm. m = 3 kg h = 0.4 m Force needed is equal to Fg W = mgh = 3(9.8)(0.4) = 12 J

Forces acting: Ff Fa Fg and FN FN does NO work. 6.3 A push of 200 N moves a 100 N block up a 30inclined plane. The coefficient of kinetic friction is 0.25 and the length of the plane is 12 m. a. Find the work done by each force acting on the block. Fa = 200 N Fg = 100 N θ = 30˚ μ = 0.25 r = 12 m Forces acting: Ff Fa Fg and FN FN does NO work.

WFg = Fg r = -Fgxr = - Fgsin30˚r = - 100 sin 30˚ (12) = - 600 J Fgy Ff W Fa = Fa r = 200 (12) = 2400 J Ff = μ FN = μ Fgcos 30˚ = 0.25 (100) cos 30˚ = 21.6 N WFf = -Ff r = - 21.6 (12) = -259.2 J FN WFg = Fg r = -Fgxr = - Fgsin30˚r = - 100 sin 30˚ (12) = - 600 J Fgy θ Ff Fg Fgx

Net work: ΣW = 2400 - 259.2 - 600 = 1540.8 J FN The resultant force: b. Show that the net work done by these forces is the same as the work of the resultant force. Net work: ΣW = 2400 - 259.2 - 600 = 1540.8 J FN The resultant force: ΣFx = Fa - Ff - Fgx = 200 - 21.6 - 50 = 128.4 N WF = FR . r = 128.4 (12) = 1540.8 J Fgy θ Ff Fg Fgx

ENERGY Energy is that which can be converted into work. When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world.

In mechanics we are concerned with two kinds of energy:   KINETIC ENERGY: K.E, energy possessed by a body by virtue of its motion. Units: Joules (J) POTENTIAL ENERGY: PE, energy possessed by a system by virtue of position or condition. PE = m g h Units: Joules (J)

WORK-ENERGY PRINCIPLE: The work of a resultant external force on a body is equal to the change in kinetic energy of the body. W =  KE Units: Joules (J)

6.4 Find the kinetic energy of a 3200 N automobile traveling at 75 km/h? Fg = 3200 N v = 20.8 m/s m = Fg/g = 326.5 kg KE = ½ mv2 = ½ (326.5) (20.8)2 = 7.06 x104 J

6.5 What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of 12 cm? m = 0.016 kg x = 0.12 m vo = 260 m/s vf = 0 m/s W = ΔKE = - 4506.7 N

PE = mgh = 0.25 (9.8) (0.2) = 0.49 J PE = mgh = 0.25 (9.8) (1.2) 6.6 A 250 g object is held 200 mm above a workbench that is 1 m above the floor. Find the potential energy relative to a. the bench top PE = mgh = 0.25 (9.8) (0.2) = 0.49 J m = 0.25 kg h = 0.2 m b. the floor h = 1.2 m PE = mgh = 0.25 (9.8) (1.2) = 2.94 J

W = PE

CONSERVATIVE AND NON-CONSERVATIVE FORCES The work done by a conservative force depends only on the initial and final position of the object acted upon. An example of a conservative force is gravity. The work done equals the change in potential energy and depends only on the initial and final positions above the ground and NOT on the path taken.

Friction is a non-conservative force and the work done in moving an object against a non-conservative force depends on the path. For example, the work done in sliding a box of books against friction from one end of a room to the other depends on the path taken.

LAW OF CONSERVATION OF ENERGY The law of conservation of energy states that: "Energy is neither created nor destroyed." Energy can be transformed from one kind to another, but the total amount remains constant. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system.

6. 7 A 40 kg ball is pulled to one side until it is 1 6.7 A 40 kg ball is pulled to one side until it is 1.6 m above its lowest point. What will its velocity be as it passes through its lowest point? m = 40 kg h = 1.6 m PE = K mgh = ½ mv2 = 5.6 m/s

ΔPE = ΔK ΔPE = m2gh – m1gh = 0.7 ( 9.8) (1.2) - 0.8 (9.8) (1.2) 6.8 In an Atwood machine the two masses are 800 g and 700 g. The system is released from rest. How fast is the 800-g mass moving after it has fallen 120 cm? ΔPE = ΔK ΔPE = m2gh – m1gh = 0.7 ( 9.8) (1.2) - 0.8 (9.8) (1.2) = - 1.17 J m1 = 0.8 kg m2 = 0.7 kg h = 1.2 m loss of PE = gain in K = 1.17 J = 1.25 m/s

At point A Energy: PEA + KA At point B PEA + KA = KB = 4.4 m/s 6.9 If friction forces are negligible and the bead has a speed of 200 cm/s at point A, a. What will be its speed at point B? A vA = 2 m/s hA = 0.8 m hB = 0 m C B At point A Energy: PEA + KA At point B PEA + KA = KB = 4.4 m/s

At point C PEA + KA = PEC + KC = 3.14 m/s b. What will be its speed at point C? hC = 0.5 m A At point C PEA + KA = PEC + KC C B = 3.14 m/s

GENERAL CASE   In real life applications, some of the mechanical energy is lost due to friction. The work due to non-conservative forces is given by:   WNC = Δ K + Δ PE

WFf = Δ K + Δ PE = Ff r = 0.015(9.8)(0.5-0.8) = - 0.04 J 6.10 Suppose the bead in Prob. 6.9 has a mass of 15 g and a speed of 2 m/s at A, and it stops as it reaches point C. The length of the wire from A to C is 250 cm. How large is the average friction force that opposes the motion of the bead? m = 0.015 kg r = 2.5 m vA = 2 m/s vC = 0 m/s hA = 0.8 m hC = 0.5 m WFf = Δ K + Δ PE = Ff r = 0.015(9.8)(0.5-0.8) = - 0.04 J = 0 - ½ (0.015)(2)2 = - 0.03 J = - 0.028 N

6.11 A 64 N block rests initially at the top of a 30 m plane inclined at an angle of 30. If k = 0.1. Find the final velocity at the bottom of the plane from energy considerations. Fg = 64 N m = 64/9.8 = 6.5 kg h = 30 m μ = 0.1 r = 15 m r r = 30 sin 30˚= 15 m h At the TOP PE = mgh = 64(15) = 960 J

At the BOTTOM PE = K + WFf Ff = μ FN = μ Fg cos 30˚ Fg = 64 N m = 64/9.8 = 6.5 kg h = 30 m μ = 0.1 r = 15 m At the BOTTOM PE = K + WFf Ff = μ FN = μ Fg cos 30˚ = 0.1 (64 cos 30˚ ) = 5.54 N WFf = Ff r = 5.54 (15) = 166.2 J K = PE - WFf = 960 - 166.2 = 793.8 J K = ½ mv2 = 15.6 m/s

POWER Is the rate at which work is performed. P = work/time P = W = F r = F v t t Units: J/s: watt (W)

6.12 An advertisement claims that a certain 1200-kg car can accelerate from rest to a speed of 25 m/s in a time of 8.0 s. What power must the motor produce to cause this acceleration? m = 1200 kg vo = 0 m/s vf = 25 m/s t = 8 s W = ΔK = ½ (1200)(25)2= 3.75x105 J = 47 kW

6. 13 A 0. 25 hp motor is used to lift a load at the rate of 5 cm/s 6.13 A 0.25 hp motor is used to lift a load at the rate of 5 cm/s. How great a load can it lift at this constant speed? (1 hp = 746 W) P = 0.25 hp (746 W/hp) = 186.5 W v = 0.05 m/s P = Fv = 3730 N = 380.6 kg

ELASTIC POTENTIAL ENERGY  Elastic potential energy is associated with elastic materials. The force Fp applied to a spring to stretch it or to compress it an amount x is directly proportional to x. That is: Fp = k x   Where k is a constant called the spring constant and is a measure of the stiffness of the particular spring. The spring itself exerts a force in the opposite direction: Fs = - k x Units: Newtons (N) This force is sometimes called restoring force because the spring exerts its force in the direction opposite to the displacement. This equation is known as the spring equation or Hooke’s Law.

The elastic potential energy is given by: PEs = ½ kx2 Units: Joules (J)

6.14 A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its normal length, what speed does the dart acquire? m = 0.1 kg k = 250 N/m x = 0.06 m PEs = K ½ kx2 = ½ mv2 = 3 m/s

6.15 A 0.20 kg ball is attached to a vertical spring as in the figure. The spring constant k is 28 N/m. The ball, supported initially so that the spring is neither stretched nor compressed, is released from rest. How far does the ball fall before being brought to a momentary stop by the spring? m = 0.20 kg k = 28 N/m vo = 0 m/s ho = 0.5 m PEg = PEs mgho = ½ kx2 = 0.26 m