PHY166 Fall 2005 6 – Work and Energy

Work - The concept following from the analysis of simple machines

F input Lever Pulley input F2 d output F1 2

d 2 output

FF F=mg/2 • Simple machines allow to gain in force: z achieve a greater output force with a smaller input force Load • Gaining in force one loses in distance mg

• Concept of work :

Work = force × distance is conserved, (work input) = (work output) 1 Definition of Work Work: W = F d = Fd cos θ = F •d F - force x Force components perpendicular to displacement do not produce work ! θ d - displacement Work is negative, if cos θ <0 x Unit of work: J(oule) J = N(ewton) m = kg m 2/s 2 Example: Pull

Fy F Problem:

FN m = 50 kg, F=100 N, F fr =50 N θ θ = 37°, d = 40 m Fx Work done by each force - ? Ffr Solution = ° = = ° = WG mgd cos 90 ,0 WN FN d cos 90 0 = θ = × × ° = WPull Fd cos 100 N 40 m cos37 3200 J d = ° = × × = − mg Wfr Ffr d cos 180 50 N 40 m ( - )1 2000 J

2 Power - Rate of doing work W W d P = or P = = F = Fv t t t

Car For P = const

Greater v Smaller F smaller acceleration a=F/m

Unit of power: W(att)

W = J(oule) /s = kg m 2/s 3

3 (Mechanical) Energy - Work stored in a body or ability of a body to do work Mechanical energy = Kinetic energy + Potential energy

E = Ekin + Epot mv 2 E = , E - different forms kin 2 pot Work done Increase of energy

= Illustration for the linear motion with constant acceleration (v0 0) 2 2 1 2 m(at ) mv W = F × d = ma × at = = - Kinetic energy 2 2 2

In general: = − W21 E2 E1

Work of external forces done on the way from position 1 to position 2 Equals the change of energy of the system 4 Potential Energy - Work needed to bring a system into another state quasistatically (v →0 )

Gravitational energy Elastic energy

2 F = kx

kx F k – stiffness of spring F d = h x – elongation/kompression

d = x 1 1 1 Ground level E = W = kx × x = kx 2 mg potential 2 2

kx a = 0 F = mg

E = W = Fh = mgh force potential Work

distance x 5 Conservation of Energy

In the absence of dissipation (friction) the total energy of an isolated system is conserved: ≡ = + = E Etot Epot Ekin const

Energies of the two different kinds can be transformed into each other: - potential energy can be released into kinetic energy - kinetic energy can be absorbed into potential energy

Problem At 1: m=0.5 kg, h=12 cm, v=0 Speed at 2 ? Solution E = E + E h 3 h 1 kin 1, pot 1, 1 = E = E + E v E = mgh 2 kin 2, pot 2, E = mgh 2 potential potential 2 E = E = ,0 = mv kin 1, pot 2, Ekinetic thus 2 mv 2 E = mgh = E = 2 pot 1, kin 2, 2 Oscillations! = = × 2 × = 6 v2 2gh 2 8.9 m/s 12.0 m 53.1 m/s Problem A dart of a mass 0.100 kg is pressed against the spring of a toy dart gun. The spring with spring stiffness k=250 N/m is compressed 6.0 cm and released. If the dart detaches from the spring when the spring is reaching ist natural length (x=0) what speed does the dart acquire?

Known : m = 0.1 kg, k = 250 N/m, x1 = 6 cm = 0.06 m To find : v2 - ? Solution : The total energy of the system spring + dart is conserved State 1: Deformed spring, potential energy State 2: Flying dart, kinetic energy

2/1 x2 = (x2 ) = x2× 2/1 = x1 = x 1 mv 2 kx 2 k = ⇒ 2 = 2 ⇒ = 1 = E1 E2 kx 1 v2 x1 2 2 m m More accurately:

General analytical result x2 =| x | Plugging numbers: 250 N/m v = 06.0 m = 06.0 2500 = 06.0 ×50 = 3 m/s 2 1.0 kg N/m kg m/s 2 / m Check units separately: m = m = m 1/s 2 = m/s, OK 7 kg kg Homework:

Giancoli Chapter 6, Problems 1, 5, 9, 15, 27, 39, 59

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