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 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 A box rests on an 30 deg inclined surface, with Static Coeff = 0.2 and Kinetic Coeff = 0.1 . A girl pushes on the 10Kg 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 Prove using Vf2-Vo2 = 2ad Work done by a Force = Change in Kinetic Energy Prove using Vf2-Vo2 = 2ad

If KE is lost where does it go? If KE is gained where does it come from?

change in vertical height Work done by gravity end dist dist∥ start 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 in PEgrav = -Workgrav Workgrav = -change in PEgrav

If gravity is the only force doing work…. Work-Energy theorem: - change in mgh = change in ½ mv2 0 = change in mgh + change in ½ mv2 change in (mgh + ½ mv2) = 0 mgh + ½ mv2 = constant

Conservation of energy mgh + ½ mv2 = constant Gravitational Potential energy Kinetic energy If gravity is the only force that does work: PE + KE = constant Energy is conserved

Free fall height t = 0s 80m V0 = 0 75m t = 1s V1 = 10m/s 60m t = 2s

m=1kg free falls from 80m mgh ½ mv2 sum t = 0s V0 = 0 h0=80m 800J 0 750J 50J V1 = 10m/s; h1=75m 800J t = 2s V2 = 20m/s; h2=60m 600J 200J 800J t = 3s V3 = 30m/s; h3=35m 350J 450J 800J t = 4s V4 = 40m/s; h4=0 0 800J 800J

T is always ┴ to the motion pendulum T W=mg Two forces: T and W T is always ┴ to the motion (& does no work)

Pendulum conserves energy E=mghmax E=mghmax hmax E=1/2 m(vmax)2

Roller coaster

Spring - Hooke’s Law

Work done by spring = - change in ½ kx2 Work done by a spring Relaxed Position F=0 x F When you compress the spring (+ work done; spring does -work) Work done by spring = - change in ½ kx2

Spring Potential Energy Workspring = -change in ½ kx2 This is the: “Spring’s Potential Energy” (or PEspring) Workspring = -change in PEspring 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) Frictional Surface H

If spring is the only force doing work…. Work-energy theorem: -change in ½ kx2 = change in ½ mv2 0 = change in ½ kx2 + change in ½ mv2 change in ( ½ kx2 + ½ mv2) = 0 ½ kx2 + ½ mv2 = constant

Conservation of energy springs & gravity mgh + ½ kx2 + ½ mv2 = constant Gravitational potential energy spring potential energy 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 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 a=g/5 (1) During the fall through a distance d = 12 m, what is the work Wg done on the cab by the gravitational force ? 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?