Work done and KE Work done Workdone by
Work done and KE
Work done • Workdone by a force is defined as Force Displacement = Fd Cosθ
A box rests on a horizontal, frictionless surface. A girl pushes on the box with a force of 18 N to the right and a boy pushes on the box with a force of 12 N to the left. The box moves 4. 0 m to the right. Find the work done by (a) the girl, (b) the boy, and (c) the net force.
A box rests on an 30 deg inclined surface, with Static Coeff = 0. 2 and Kinetic Coeff = 0. 1. A girl pushes on the 10 Kg box with a force of 100 N upwards. The box moves 1. 0 m on the incline. Find the work done by (a) the girl, (b) the gravity, and (c) the friction (d) net Force.
What do you think? • If a Force increases velocity of an object , would you call the Wok done by this force as POSITIVE or NEGATIVE ? Why?
Work-Energy Theorem Work done by a Force = Change in Kinetic Energy Prove using Vf 2 -Vo 2 = 2 ad
If KE is lost where does it go? If KE is gained where does it come from?
Work done by gravity end dist start dist∥ change in vertical height W=mg Work = F x dist∥ = -mg x Change in height =- Change in mg h
Gravitational Potential Energy Workgrav = -Change in mgh This is called: “Gravitational Potential Energy” (or PEgrav) Change Workgravin=PE -change in PEgrav = -Work grav
If gravity is the only force doing work…. Work-Energy theorem: - change in mgh = change in ½ mv 2 0 = change in mgh + change in ½ mv 2 change in (mgh + ½ mv 2) = 0 mgh + ½ 2 mv = constant
Conservation of energy mgh + ½ Gravitational Potential energy 2 mv = constant Kinetic energy If gravity is the only force that does work: PE + KE = constant Energy is conserved
Free fall height t = 0 s V 0 = 0 t = 1 s 80 m 75 m V 1 = 10 m/s 60 m t = 2 s V 2 = 20 m/s t = 3 s 35 m V 3 = 30 m/s t = 4 s V 4 = 40 m/s 0 m
m=1 kg free falls from 80 m t = 0 s V 0 = 0 h 0=80 m mgh 800 J ½ mv 2 sum 0 800 J 50 J 800 J t = 1 s V 1 = 10 m/s; h 1=75 m 750 J t = 2 s V 2 = 20 m/s; h 2=60 m 600 J 200 J 800 J 350 J 450 J 800 J t = 3 s V 3 = 30 m/s; h 3=35 m t = 4 s V 4 = 40 m/s; h 4=0 0 800 J
pendulum T W=mg Two forces: T and W T is always to the motion (& does no work) ┴
Pendulum conserves energy E=mghmax E=1/2 m(vmax)2
Roller coaster
Spring - Hooke’s Law
Work done by a spring Relaxed Position F=0 F x When you compress the spring (+ work done; spring does -work) Work done by spring = - change in ½ kx 2
Spring Potential Energy Workspring = -change in 2 ½ kx This is the: “Spring’s Potential Energy” (or PEspring) Workspring = -change in PEspring = -Workspring
What are the graphical models and the mathematical models that are used to represent the energy of a system? Frictionless surface ( hypothetical) H Frictional Surface
If spring is the only force doing work…. Work-energy theorem: -change in ½ kx 2 = change in ½ mv 2 0 = change in ½ kx 2 + change in ½ mv 2 change in ( ½ kx 2 + ½ mv 2) = 0 ½ 2 kx +½ 2 mv = constant
Conservation of energy springs & gravity mgh + 2 ½ kx + 2 ½ mv Gravitational spring potential energy = constant Kinetic energy If elastic force & gravity are the only force doing work: PEgrav + PEspring + KE = constant Energy is conserved
Summary • Net Work Done on an object = KE final – KE initial That it is + when the system gains KE energy And it is - when it loses KE energy
An elevator cab of mass m = 500 kg is descending with speed vi = 4. 0 m/s when its supporting cable begins to slip, allowing it to fall with constant acceleration (1) During the falla=g/5 through a distance d = 12 m, what is the work Wg done on the cab by the gravitational force ? (2) During the 12 m fall, what is the work WT done on the cab by the upward pull of the elevator cable?
(3) What is the net work W done on the cab during the fall? (4) What is the cab's kinetic energy at the end of the 12 m fall?
- Slides: 29