Work and Energy Physics Chapter 5

Chapter 5 Objectives Recognize the difference between the scientific and ordinary definitions of works Define work by relating it to force and displacement Calculate the net work done with many forces are applied to an object

Work How do you define work?

Definition of Work Work (W) is done on an object when a force (F) causes a displacement (d) of the object Work is done on an object only if it moves in the direction of the force W = F d Force units (N)  distance units (m) N•m are also called joules (J) W =  F ds

Work 2 categories of Work Work done against another force Lift something – exerted against force of gravity Work done to change the speed of something Stopping a car or speeding it up How would you calculate the work in this case? What is the component of F in the direction of d? Only the component of the force that is in the direction of the objects displacement does work! F cos  W = F cos  d

Energy Indentify several forms of energy Calculate kinetic energy for an object Apply the work-kinetic energy theorem to solve problems Distinguish between kinetic & potential energy Classify different types of potential energy Calculate the potential energy associated with an object’s position

Energy Energy is “something” that enables an object to do work What is that something? Energy is associated with heat, light, electricity, mechanical motion, sound, and the nature of a chemical reaction Mechanical energy - Energy due to the position or the movement of something Mechanical energy has 2 forms Kinetic Energy (KE) – energy of motion Potential Energy (PE) – Stored energy

Kinetic Energy Kinetic Energy depends on speed and mass Units – Joules (used for all forms of energy) kg•m2/s2 or N•m or J

Kinetic Example Problem A 100. Kg Linebacker moves at 8.90 m/s (runs 100m in 11.2 secs). How much KE does the linebacker have? How fast does a 54.5 Kg wide receiver have to run to have the same KE as the linebacker? Given: Linebacker VL = 8.90 m/s ML = 100. kg KEL = ? Wide Receiver Vw = ? Mw = 54.5 kg KEW = KEL ½ MwVw2 = KEL Vw2 = 2KEL/ Mw Vw2 = 2 ( 3.96x103 J) / (54.5 kg) Vw = 12.1 m/s Soln: KEL = ½ ML VL2 = ½ (100. Kg) (8.90 m/s)2 KEL = 3.96 x 103 J note: wide receiver would have to run 100m in 8.26 secs

Work-Kinetic Energy Theorem The net work (Wnet) done on an object is equal to the change in the KE of the object Wnet =  KE = ½ mvf2 - ½ mvi2

Example problem P. 168 #2 Wnet =  Kef -  Kei W=Fnetd Ff = 950 N Given: m = 2.0 x 103 Kg vi = 0 Vf = 2.0 m/s d = ? FA= 1140 N Fnetd = ½ mVf2 d = mVf2 2Fnet d = (2.0 x 103 Kg) (2.0 m/s)2 2 (1140N – 950N) d = 21 m Soln: Wnet =  Kef -  Kei Wnet = ½ mvf2 - ½ mvi2 W=Fnetd FNetd = ½ mVf2 - ½ mVi2

gravitational PE = mass  free-fall acceleration  height Potential Energy Potential Energy is the energy associated with an object because of the position, shape, or condition of the object. Stored energy Chemical energy Object due to its position (Gravitational, ex: rock on hilltop) Elastic potential energy Gravitational potential energy is the potential energy stored in the gravitational fields of interacting bodies. depends on height from a zero level (zero level is arbitrary) PEg = mgh gravitational PE = mass  free-fall acceleration  height

Potential Energy Elastic potential energy is the energy available for use when a deformed elastic object returns to its original configuration. The symbol k is called the spring constant, a parameter that measures the spring’s resistance to being compressed or stretched.

Potential Energy Elastic Potential energy The energy available for use in deformed elastic objects Rubber bands, springs in trampolines, pole-vault poles, muscles For springs, the distance compressed or stretched = x

Sample Problem: Elastic potential energy A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has stretched a length of 44.0 m. Disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling? Given: m = 70.0 kg k = 71.8 N/m g = 9.81 m/s2 h = 50.0 m – 44.0 m = 6.0 m x = 44.0 m – 15.0 m = 29.0 m PE = 0 J at river level PEtot = ?

Sample Problem: Elastic potential energy The zero level for PEg is chosen to be at the surface of the water. h = 50.0 m – 44.0 m = 6.0 m h= 6.0 m The total Potential Energy is the sum of PEg and PEelastic PEtot = PEg + PEelastic PEg = mgh = (70.0 kg)(9.81m/s2)(6.0m) = 4.1x103 J PEelastic = ½ kx2 = ½ (71.8 N/m)(29.0m)2 = 3.02x104J PEtot = 4.1x103J + 3.02x104J PEtot = 3.43x104J

Conservation of Energy Energy cannot be created or destroyed, but only changed from one form to another. It transforms without net loss or gain When we say that some is conserved, we mean that it remains constant 2 types of Energy Mechanical Energy –kinetic, gravitational Potential, elastic potential, and chemical potential energy 2. Nonmechanical energy – Nuclear, chemical, internal, and electrical

Mechanical Energy MEi = MEf ME = KE + ∑PE Mechanical Energy is the sum of KE and PE present in a situation ME = KE + ∑PE Mechanical Energy is also conserved MEi = MEf initial mechanical energy = final mechanical energy (in the absence of friction)

Mechanical Energy example p. 177 #2 Initial hi = 10.0m KEi = 0 PEi = mghi Midpoint hm = 5.0m KEm = ½ mVm2 PEm = mghm Final hf = 0 KEf = ½ mVf2 PEf = 0 Given: Soln: @ midpoint (hm = 5.0) MEi = MEm PEi + KEi = PEm + KEm mghi = mghm + ½ mVm2 Vm2 = 2ghi – 2ghm = 2g(hi-hm) Vm = 9.9 m/s @ Final MEi = MEf PEi + KEi = PEf + KEf mghi = ½ mVf2 Vf2 = 2ghi Vf = 14.0 m/s

power = work ÷ time interval Power is a quantity that measures the rate at which work is done or energy is transformed How fast you do the job Rate of energy transfer P = W/∆t power = work ÷ time interval Units for power – Watts (W) = J/s Horsepower (hp) another unit for power 1hp = 746 watts

Power Machines with different power ratings do the same amount of work in different time intervals An alternate equation for power in terms of force and speed is P = Fv power = force  speed