PHYSICS 149: Lecture 17

• Chapter 6: Conservation of Energy – 6. 7 Elastic Potential Energy – 6.8 Power

• Chapter 7: Linear Momentum – 7.1 A Vector Conservation Law – 7.2 Momentum

Lecture 17 Purdue University, Physics 149 1 ILQ 1

A force of 5 N is applied to the end of a spring, and it stretches 10 cm. How much "farther" will it stretch if an "additional" 2.5 N of force are applied?

A) 2.5 cm B) 5 cm C) 10 cm D) 15 cm

Lecture 17 Purdue University, Physics 149 2 ILQ 2

A mass is attached to the bottom of a vertical sppgring. This causes the s pgpring to stretch and the mass to move downward. Does the potential energy of the spring increase or decrease? Does the gravitational potential energy of the mass increase of decrease?

A) PE of spring decreases; PE of mass increases B) both decrease C) bot h increase D) PE of spring increases; PE of mass decreases

Lecture 17 Purdue University, Physics 149 3 ILQ 3

A simple catapult, consisting of a leather pouch attached to rubber bands tied to two forks of a wooden Y, has a spring constant k and is used to shoot a pebble horizontally. When the catapult is stretched by a distance d, it gives the pebble a speed v.

What speed does it give the same pebble when it is stretched to a distance 4d?

A) 16 v B) 4 v C) 4 sqrt(4) v D) 64 v E) sqrt(4) v Lecture 17 Purdue University, Physics 149 4 Energy and Work

• Work: Transfer of Energy by Force

WF = |F| |s| cosθ θ

• Kinetic Energy (Energy of Motion) K = 1/2 mv2

• Work-Energy Theorem:

ΣWNC = ΔK+K + ΔU • Gravitational Potential Energy: mm12 Ugrav = mgy = -Wgrav UG=− Lecture 17 Purdue University, Physics 149r 5 Hooke’s Law

• The deformation–change in size or shape–of the object is proportional to the magnitude of the force that causes the deformation.

– Magnitude: k|x| – Direction: Whether the spring is compressed or stretched, the Hooke’s force always points toward its relaxation position (x=0).

– Spring constant k • A characteristic of a spring x=0 • Unit: N/m |Fx|=k|x| to left

Lecture 17 Purdue University, Physics 149 6 Work by Variable Force

•W = Fx Δx Force – Work is area under F vs x plot Work – Spring: F = k x 2 Distance • Area = ½ k x =Wspring Work is the area under the F vs x plot

Force

Work Distance

Lecture 17 Purdue University, Physics 149 7 Work and Potential Energy of a Spring • The force of a spring F = -kx • The work

xi = 0 x f = x 1 2 1 2 1 2 W = − kx f + kxi W = − kx 2 2 2

• The potential energy

xi = 0 x f = x 1 1 ΔU = −W = kx 2 − kx 2 1 spring 2 f 2 i U = kx 2 2

Lecture 17 Purdue University, Physics 149 Lecture 15 page 8 ILQ

• You compress a spring by a distance x and store 10 J of energy. How much energy is stored if you compress the spring a distance 2x?

a) 5 J b) 20 J c) 30 J d) 40 J

Lecture 17 Purdue University, Physics 149 9 Work Done by an Ideal Spring • The work done by a spring as its movable end moves from

equilibrium (xi=0) to the final position xf is

x f G G Wspring = FHooke ⋅ds ∫xi =0 beyond the scope x = f (−kx) dx of this class ∫x =0 i 1 = − kx2 2 f Å independent of path ! • The work done by the spring is the

(ti)(negative) area un dthder the Fx(x)h) graph. Area = ½⋅base⋅height • The work done by an ideal spring depends only on the initial and final positions of the moveable end (independent of the path taken). Æ Hooke’ s force is a conservative force . Æ Potential energy may be defined.

Lecture 17 Purdue University, Physics 149 10 Elastic Potential Energy

• The work done by an ideal spring is independent of path (see previous slide) Æ It means that the spring force is a conservative force Æ Thus, we may d efi ne pot enti al energy f or th e spr ing force.

ΔU = U (x f ) −U (xi ) ≡ −WHooke 1 2 1 2 U (x f ) = U (xi ) + kx f − kxi x f G G 2 2 beyond the scope = − FHooke ⋅ds ∫x i of this class x = − f (−kx) dx ∫xi If we let U (xi ) ≡ 0 at xi = 0, 1 1 1 = kx2 − kx2 U (x ) = kx2 2 f 2 i f 2 f

arbitrary choice Lecture 17 Purdue University, Physics 149 11 ILQ • All springs and masses are identical. (Gravity acts down). – Which of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position ? A) 1 B) 2 C) same

(1) (2) Lecture 17 Purdue University, Physics 149 12 ILQ: Solution

• The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg)

0 d 2d

(1) (2) Lecture 17 Purdue University, Physics 149 13 ILQ: Solution 1 • The potential energy stored in (1) is 2 ⋅=kd22 kd 2 1 2 • The potential energy stored in (2) is k2d( ) = 2kd2 2

The spr ing P. E. is twice as big in (2) ! 0 d 2d

(1) (2)

Lecture 17 Purdue University, Physics 149 14 Vertical Springs • A spring is hung vertically. Its relaxed position is at y = 0 (a). When a mass m is hung from (a) (b) its end, the new equilibrium j k position is ye (b). z Recall that the force of a spring is Fs = -kx. In case (b) y = 0 Fs = mg and x = ye:

y = ye -kye - mg = 0 (ye < 0) m

-ky mg = -kye mg e

(ok since ye is a negative number) Lecture 17 Purdue University, Physics 149 15 Vertical Springs • The pot enti al energy of th e spring-mass system is: (a) (b) j 1 2 U =++ky mgy C k 2 but mg = -ky e y = 0 1 U = ky 2 − ky y +C 2 e y = ye m choose C to make U=0 at y=yy = ye:

mg -kye 1 2 2 1 0 = ky −ky +C C = kky 2 2 e e 2 e Lecture 17 Purdue University, Physics 149 16 Vertical Springs •So:

11 (a) (b) Uk=−+y2 kyy ky 2 22ee j k 1 2 2 =+−ky( yee 2 yy) 2 y = 0 which can be written: y = ye m 1 2 U = k(y − y ) e -ky 2 mg e

Lecture 17 Purdue University, Physics 149 17 Vertical Springs

1 2 U = k(y −ye ) (a) (b) 2 j • So if we define a new y′ k coordinate system such that y′ = 0 is at the equilibrium position, (

y′ = y - ye ) then we get the simple result: y′ = 0 m 1 U = ky ′ 2 2

Lecture 17 Purdue University, Physics 149 18 Vertical Springs • If we choose y = 0 to be at the eqqpuilibrium position of the mass (()a) (()b) hanging on the spring, we can define j the potential in the simple form. k 1 U = ky 2 2 • Notice that g does not appear in this expression!! y = 0 m – BhBy choos ing our coor ditdinates an d cons tttants cleverly, we can hide the effects of gravity.

Lecture 17 Purdue University, Physics 149 19 ILQ: Energy Conservation • In (1) a mass is hanging from a spring. In (2) an identical mass is held at the height of the end o f th e same spri ng in its re laxe d pos ition. – Which correctly describes the relation of the potential energies of the two cases?

(a) U1 > U2 (b) U1 < U2 (c) U1 = U2

case 1 case 2

d

Lecture 17 Purdue University, Physics 149 20 ILQ: Solution • In case 1 , it i s s imp les t to c hoose the mass to have zero total potential energy (sum of spring and gravitational potential energies) at its equilibrium position. 1 • In case 2 the total potential energy is then U = kd 2 2 2

relaxed

y = d d y = 0, U1 = 0

The answer is (b) U1 < U2. Lecture 17 Purdue University, Physics 149 21 Power (Rate of Work)

• P = W / Δt ΔW P = – Units: Joules/Second = Watt Δt . F •W = F Δr = F Δr cosθ Δr = F (v Δt) cosθ • P=FvP = F v cosθ v • How much power does it take for a (70 kg) student to run up the stairs (5 meters) in 7 seconds? P = W / t = m g h / t = (70 kg) (9. 8 m/s2) (5 m) / 7 s = 490 J/s or 490 Watts Lecture 17 Purdue University, Physics 149 22 Example

• Lars, of mass 82.4 kg, has been working out and can do work for about 2.0 min at the rate of 746 W. How long will it take him to climb three flights of stairs, a vertical height of 12.0m?

– As Lars climbs the stairs, he increases his gravitational potential energy. – The rate of potential energy increase must be equal to the rate he does work.

ΔE ΔU mgΔy P = = = av Δt Δt Δt mgΔy (82.4kg)(9.80m / s2 )(12.0m) ⇒ Δt = = =13.0s Pav 746W

Lecture 17 Purdue University, Physics 149 23 Power y •A 2000 kg trolley is pulled up x a 30 degree hill at 20 mi/hr v T by a w inc h a t the top o f the winch hill. How much power is the winch providing? θ mg • The power is P = F.v = T.v • Since the trolley is not accelerating, the net force on it must be zero. In the x direction: – T - mg sin θ = 0 – T = mg sin θ

Lecture 17 Purdue University, Physics 149 24 Power y • P= T.v = Tv x v since T is parallel to v T winch •SoP = mgv sin θ θ v = 20 mi/hr = 8.94 m/s mg g = 9.81 m/s2 m = 2000 kg sin θ = sin(30o) = 0.5 and P = (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5) = 87,700 W

Lecture 17 Purdue University, Physics 149 25 Power • Power is the rate at which energy is transferred, or equivalently, the rate at which work is done (that is, work per a time interval).

– Average Power: the average rate of energy conversion

– Instantaneous Power: the instantaneous rate at which a force F does work when the object it acts on moves with velocity v

W FΔr cosθ P = = = Fvcosθ Δt Δt

• Power is a scalar quantity. • Units: W, J/s, etc. – Unit conversion: 1 W = 1 J/s – Note that kWh (kilowatt-hour) is a unit of energy, not power. • Power is denoted by P. Lecture 17 Purdue University, Physics 149 26 ILQ

• What power must an engine have if it is to be used to raise a 25 kg load 10 m in 4 seconds?

a) 25 W b) 625 W c) 1000 W d) 2500 W

Lecture 17 Purdue University, Physics 149 27 Example: The Dart Gun

In this case, there are three forces acting on the dart. But, the directions of gravity and normal forces are perpendicular to the displacement of the dart, so the work done by the two forces are zero. Hooke force (a conservative force) is the only force which does work, so the mechanical energy is conserved.

Emech = Ki + Ui = Kf + Uf = const (b/c Wnc =0)= 0)

Ki = 0 (b/c vi = 0) 2 Ui = ½kxi 2 Kf = ½mvf Uf = 0 (b/c xf = 0)

2 2 Æ 0 + ½kxi = ½mvf + 0 Thus, vf = sqrt(k/m)⋅|xi| = 11 m/s Lecture 17 Purdue University, Physics 149 28 Key Ideas • Work-Energy – Σ F = m a multiply both sides by d – Σ F d = m a d (note: a d = ½ Δv2) – Σ F d = ½ m Δv2 – Σ W = ΔK DfiDefine Wor k an dKitiEd Kinetic Energy

•Impulse-Momentum – Σ F = m a multiply both sides by Δt – Σ F Δt = m a Δt (note: a Δt = Δv) – Σ F Δt = m Δv – Σ I = Δp Define Impulse and Momentum

Lecture 17 Purdue University, Physics 149 29 Momentum is Conserved

• Momentum is “Conserved” meaning it can not be created nor destroyed – Can be transferred

• Total Momentum does not change with time

• Momentum is a VECTOR 3 Conservation Laws in one!

Lecture 17 Purdue University, Physics 149 30 A Vector Conservation Law

• When a vector quantity is conserved in an interaction, both its magnitude and direction are unchanged (or equivalently, all components are unchangg)ed).

Lecture 17 Purdue University, Physics 149 31 Example: Momentum

• What is the momentum of an automobile (weight = 9800 N) when it is moving at 35 m/s to the south?

– W = mg Æ m = W/g = (9800 N) / (9.8 m/s2) = 1000 kg

– p = mv = (1000 kg ) ⋅ (35 m /s sou th) = 35,000 kg⋅m/s south

Don’t forget that momentum p is a vector! (We need both its magnitude and direction.)

Lecture 17 Purdue University, Physics 149 32 Pushing Off… • Fred and Jane are on skates facing each other. Jane then pushes Fred with force F

– N2L Fred: FJF = mFred a a = Δv/Δt

ΔvFred = a Δt

= (FJF/mFdFred) Δt

mFred ΔvFred = FJF Δt

– N2L Jane: FFJ = mJane a

ΔvJane = a Δt

= (F/mJane) Δt

mJane ΔvJane = FJF Δt – N3L: For every action, there is an equal and opposite

reaction. FFJ=-FJF

mFred ΔvFred = -mJane ΔvJane Lecture 17 Purdue University, Physics 149 33 Pushing Off… Fred (75 kg) and Jane (50 kg) are on skates facing each other. Jane then pushes Fred with a constant force F = 45 N for a time Δt = 3 seconds. Who will be moving fastest at the end of the push? A) Fred B) Same C) Jane

Fred Jane F = +45 N (positive direct.) F = -45 N Newton’s 3rd law I = +45 × 3 Ns = 135 Ns I = -45 × 3 Ns = -135 Ns I = Δp I = Δp

= mvf –mvi = mvf –mvi

I/m = vf -vi I/m = vf -vi

vf = 135 N-s / 75 kg vf = -135 N-s / 50 kg = 181.8 m/ s = -272.7 m/ s

Note: Pfred + Pjane = (1.8) 75 + (-2.7) 50 = 0! Lecture 17 Purdue University, Physics 149 34