2.3.1outline what is meant by work. 2.3.2determine the work done by a non-constant force by...

2.3.1 Outline what is meant by work.

2.3.2 Determine the work done by a non-constant force by interpreting a force-displacement graph.

2.3.3 Solve problems involving the work done by a force.

Topic 2: Mechanics2.3 Work, energy and power

Outline what is meant by work.

Dynamics is the study of forces as they are applied to bodies, and how the bodies respond to those forces.

Generally, we create free body diagrams to solve the problems via Newton’s second law.

A weakness of the second law is that we have to know all of the applied forces in order to solve the problem.

Sometimes, forces are hard to get a handle on.In other words, as the following example will show, Newton’s second law is just too hard to use to solve some physics problems.

In these cases, the principles of work and energy are used - the subjects of Topic 2.3.




EXAMPLE: Suppose we wish to find the speed of the ball when it reaches the bottom of the track.

Discuss the problems one might have in using free body diagrams to find that final speed.

SOLUTION:Because the slope of the track is changing, so is the relative orientation of N and W.Thus, the acceleration is not constant.Thus we can’t use the kinematic equations.Thus we can’t find v at the bottom of the track.

W NKEY


As we have stated, the principles of work and energy need to be mastered in order to solve this type of problem.

We begin by defining work.In everyday use, work is usually thought of as effort expended by a body, you, on homework, or on a job.

In physics, we define work W as force F times the displacement s, over which the force acts:

The units of work are the units of force (newtons) times distance (meters). For convenience, we call a newton-meter (N m) a joule (J) in honor of the physicist by that name.


W = Fs work done by a constant force




EXAMPLE: Find the work done by the 25-newton force F in displacing the box s = 15 meters.

SOLUTION:W = FsW = (25 n)(15 m)W = 380 n m = 380 J.

F s

FYI

The units for force can be abbreviated either (n) or (N).

FYI

The units of (n m) are joules (J). You can just keep them as (n m) if you prefer.


If the force is not parallel to the displacement the formula for work has the minor correction



W = Fs cos work done by a constant force not parallel to displacement

Where is the angle between F and s.

FYI

If F and s are parallel, = 0°.

If F and s are antiparallel, = 180°.

Fsparallel

Fs

antiparallel

Solve problems involving work.


W = Fs cos work done by a constant force not parallel to displacement

Where is the angle between F and s.EXAMPLE: Find the work done by the 25-newton force F in displacing the box s = 15 meters if the force and displacement are (a) parallel, (b) antiparallel and (c) at a 30° angle.

SOLUTION:(a) W = Fs cos = (25)(15) cos 0° = 380 J.

(b) W = (25)(15) cos 180° = -380 J.

(c) W = (25)(15) cos 30° = 320 J.FYI

Work can be negative.The F and the s are the magnitudes of F and s.

EXAMPLE: Find the work done by the brakes in bringing a 730-kg Smart Car to a rest in 80. meters if its starting speed is 32 m/s.

SOLUTION:Since f and s are antiparallel then = 180°.It is given that s = 80 m.From v2 = u2 + 2as we get

02 = 322 + 2a(80) so that a = -6.4 m s-2.

Then f = ma = 730(-6.4) = -4672 n. |f| = +4672 n.Finally, W = Fs cos = (4672)(80) cos 180° = -370000 J.

s



f


FBD Crate

EXAMPLE: A pulley system is used to raise a 100-n crate 4 m as shown. Find the work done by the tension force T if the lift occurs at constant speed.

SOLUTION:From the FBD since a = 0, T = 100 n.From the statement of the problem, s = 4 m.Since the displacement and the tension are parallel, = 0°.Thus W = Ts cos = (100)(4) cos 0° = 400 J.


T

TT

s

s

T

100

a = 0

FYI

Pulleys are used to redirect tension forces for convenience.

T


FBD Crate

EXAMPLE: A pulley system is used to raise a 100-n crate 4 m as shown. Find the work done by the tension force T if the lift occurs at constant speed.

SOLUTION:From the FBD 2T = 100 so that T = 50 n.From the statement of the problem, s = 4 m.Since the displacement and the tension are parallel, = 0°.Thus W = T(2s) cos = (50)(24) cos 0° = 400 J.


T

100

a = 0

FYI

Pulleys are also used gain mechanical advantage.M.A. = Fout/Fin = 100 n / 50 n = 2.

T

T

T

2s

s

T

T T

Determine the work done by a non-constant force by interpreting a force-displacement graph.

Consider a spring mounted to a wall as shown.If we pull the spring to the right, it resists in direct proportion to the distance it is stretched.

If we push to the left, it does the same thing.It turns out that the spring force F is given by

The minus sign gives the force the correct direction, namely, opposite the displacement s.

Since F is in (n) and s is in (m), the units for the spring constant k are (n m-1).


F

x0

F

F = -ks Hooke’s Law (the spring force)




EXAMPLE: A force vs. displacement plot for a spring is shown. Find the value of the spring constant, and find the spring force if the displacement is -65 mm.

SOLUTION:Pick a convenient point on the plot.For the point, F = -15 n and s = 30 mm = .030 m so thatF = -ks or -15 = -k(.030) or k = 500 n m-1.

F = -ks = -(500)(-6510-3) = +32.5 n.

F (n)20

0

-20-20 20-40 400

s (mm)




EXAMPLE: A force vs. displacement plot for a spring is shown. Find the work done by you if you displace the spring from 0 to 40 mm.

SOLUTION:The spring itself always does negative work because the force and the displacement are antiparallel.The force you apply will always be parallel. Thus you will do positive work according to F = ks.F = ks is plotted in red.

F (n)20

0

-20-20 20-40 400

s (mm)




EXAMPLE: A force vs. displacement plot for a spring is shown. Find the work done by you if you displace the spring from 0 to 40 mm.

SOLUTION:The area under the F vs. s graph represents the work done by that force.The area desired is from 0 to 40 mm, shown here:A = (1/2)bh = (1/2)(4010-3 m)(20 n)

= 0.4 J

F (n)20

0

-20-20 20-40 400

s (mm)

2.3.4 Outline what is meant by kinetic energy.

2.3.5 Outline what is meant by change in gravitational potential energy.

2.3.6 State the principle of conservation of energy.

2.3.7 List different forms of energy and describe examples of the transformation of energy from one form to another.


Outline what is meant by kinetic energy.

Kinetic energy EK is the energy of motion.

The bigger the speed v, the bigger EK.

The bigger the mass m, the bigger EK.

The formula for EK, justified later, is

Looking at the units for EK we have

kg(m/s)2 = kg m2 s-2

= (kg m s-2)m

In the parentheses we have a mass times an acceleration which is a newton.

Thus EK is measured in (n m), which are (J), the same unit we used for work.

As a final note, many books use the symbol K instead of EK for kinetic energy.


EK = (1/2)mv2 kinetic energy




EXAMPLE: What is the kinetic energy of a 4.0-gram NATO SS 109 bullet traveling at 950 m/s?

SOLUTION: Convert grams to kg (jump 3 decimal places left) to get m = .004 kg. Then K = (1/2)mv2

= (1/2)(.004)(950)2

= 1800 J.




EXAMPLE: What is the kinetic energy of a 220-pound NATO soldier running at 6 m/s?

SOLUTION: First convert pounds to kg: (220 lb)(1 kg/2.2 lb) = 100 kg. Then K = (1/2)mv2

= (1/2)(100)(6)2

= 1800 J.

FYI

Small and large objects can have the same EK!


It is no coincidence that work and kinetic energy have the same units. Observe the following derivation.

v2 = u2 + 2as

mv2 = m(u2 + 2as)

mv2 = mu2 + 2mas

mv2 = mu2 + 2Fs

(1/2)mv2 = (1/2)mu2 + Fs

EK,f = EK,0 + W

EK,f - EK,0 = W

∆EK = W

This is called the Work-Kinetic Energy theorem.





This formula says that the change in kinetic energy of an object equals the work done on it.


∆EK = W work-kinetic energy theorem

EXAMPLE: Use energy to find the work done by the brakes in bringing a 730-kg Smart Car to a rest in 80. meters if its starting speed is 32 m/s.

SOLUTION:EK,f = (1/2)mv2 = (1/2)(730)(02) = 0 J.

EK,0 = (1/2)mu2 = (1/2)(730)(322) = 370000 J.

∆EK = EK,f - EK,0 = 0 – 370000 = -370000 J

W = ∆EK = -370000 J.

sf


One of the many derivations the IBO requires you to know expresses kinetic energy in terms of momentum instead of velocity.

p = mv

p2 = m2v2

p2/m = mv2

(1/2)p2/m = (1/2)mv2

EK = p2/(2m)


EK = (1/2)mv2 kinetic energyEK = p2/(2m)

FBD Ball

Outline what is meant by change in gravitational potential energy.

Consider a bowling ball resting on the floor.If we let go of it, it just stays put.If on the other hand we raise it to a height ∆h and then let it go, it will fall and speed up, gaining kinetic energy as it falls.

Since the lift constitutes work against gravity (the weight of the ball) we have

W = Fs cos

W = mg∆h cos 0° = mg∆h.

We call this energy due to the position of a weight gravitational potential energy.


F

mg

a = 0

∆EP = mg∆h change in gravitational potential energy

EXAMPLE: Consider a crane which lifts a 2000-kg weight 18 m above its original resting place. What is the change in gravitational potential energy of the weight and how much work did the crane do on the weight?

SOLUTION:The change in gravitational potential energy is just∆EP = mg∆h = 2000(10)(18) = 360000 J.

The crane must have done 360000 J of work if the acceleration was zero.

Outline what is meant by change in gravitational potential energy.


∆EP = mg∆h change in gravitational potential energy

EXAMPLE: Consider a crane which lifts a 2000-kg weight 18 m above its original resting place. If the cable breaks at the top find the speed and kinetic energy of the mass at the instant it reaches its starting point.

SOLUTION:v2 = u2 + 2asv2 = 02 + 2(-10)(-18) = 360v = 19 m s-1.

EK = (1/2)mv2

= (1/2)(2000)(192) = 360000 J.

State the principle of conservation of energy.


EXAMPLE: Consider a crane which lifts a 2000-kg weight 18 m above its original resting place. If the cable breaks at the top find its change in kinetic energy and change in potential energy the instant it reaches the ground.

SOLUTION:EK,f = (1/2)mv2

= (1/2)(2000)(192) = 360000 J.

EK,0 = (1/2)(2000)(02) = 0 J.

∆EK = 360000 – 0 = 360000 J.

∆EP = mg∆h

= (2000)(10)(-18) = -360000 J.



EXAMPLE: Consider a crane which lifts a 2000-kg weight 18 m above its original resting place. If the cable breaks at the top find the sum of the change in kinetic and the change in potential energies the instant it reaches the ground.

SOLUTION:From the previous slide ∆EK = 360000 J and

∆EP = -360000 J so

∆EK + ∆EP = 360000 + -360000

∆EK + ∆EP = 0 J.




We see that ∆EK + ∆EP = 0 from the previous example.

In general, if there is no friction or drag to remove energy from a system

The above formula is known as the statement of the conservation of mechanical energy.

Essentially, what it means is that if the kinetic energy changes (say it decreases), then the potential energy changes (it will increase) in such a way that the total change is zero!

Another way to put it is “the total energy of a system never changes.”


∆EK + ∆EP = 0 conservation of energy

In the absence of friction and drag



EXAMPLE: Find the speed of the 2-kg ball when it reaches the bottom of the 20-m tall track.

SOLUTION:Use conservation of energy to find EK,f and v.

∆EK + ∆EP = 0

(1/2)mv2 - (1/2)mu2 + mg∆h = 0

(1/2)(2)v2 - (1/2)(2)02 + 2(10)(-20) = 0

v2 = 400

v = 20 m s-1.



∆h

List different forms of energy and describe examples of the transformation of energy from one form to another.

We have talked about kinetic energy (motion).We have talked about potential energy (position).We have chemical energy and nuclear energy.We have electrical energy and magnetic energy.We have sound energy and light energy.And we also have heat energy.In mechanics we only have to worry about the highlighted energy forms.

And we only worry about heat if there is friction or drag.




EXAMPLE: Suppose the speed of the 2-kg ball is 15 m s-1 when it reaches the bottom of the 20-m tall track. What is the loss of mechanical energy and where did it go?

SOLUTION: ∆EK + ∆EP = loss/gain

(1/2)mv2 - (1/2)mu2 + mg∆h = loss/gain

(1/2)(2)152 - (1/2)(2)02 + 2(10)(-20) = loss/gain

-175 J = loss/gain



∆h

The system lost 175 J as drag and friction heat.

2.3.8 Distinguish between elastic and inelastic collisions.


Distinguish between elastic and inelastic collisions.

In an elastic collision, kinetic energy is conserved (it does not change). Thus EK,f

= EK,0.

In an inelastic collision, kinetic energy is not conserved (it does change). Thus EK,f

≠ EK,0.

In a completely inelastic collision the colliding bodies stick together.




EXAMPLE: Two train cars having equal masses of 750 kg and velocities u1

= 10. m s-1 and u2 = 5.0 m s-1 collide and hitch together. What is their final speed?

SOLUTION:Use conservation of linear momentum.If Fnet = 0 then p = CONST, so that

p1,0 + p2,0 = p1,f + p2,f

mu1 + mu2 = mv + mv

m(u1 + u2) = 2mv

10 + 5 = 2v or v = 7.5 m s-1.

u1 u2v v



EXAMPLE: Two train cars having equal masses of 750 kg and velocities u1

= 10. m s-1 and u2 = 5.0 m

s-1 collide and hitch together. Identify the collision type/s that occur/s.

SOLUTION:Since the cars stick together we have a completely inelastic collision.From the previous slide we have v = 7.5 so that∆EK = (1/2)(2m)v2 – [ (1/2)mu1

2 + (1/2)mu22 ]

= (1/2)(2750)7.52 –(1/2)750102 -(1/2)75052

= -4700 J.

u1 u2v v

Inelastic collision since ∆EK ≠ 0.

2.3.9 Define power.

2.3.10 Define and apply the concept of efficiency.


Define power.

Power is the rate of energy usage and so has the equation

From the formula we see that power has the units of energy (J) per time (s) or (J s-1) which are known as watts (W).


P = E/t power

EXAMPLE: How much energy does a 100.-W bulb consume in one day?

SOLUTION:From P = E/t we get E = Pt so thatE = (100 J/s)(24 h)(3600 s/h)E = 8640000 J!Don’t leave lights on in unoccupied rooms.

Define and apply the concept of efficiency.

Efficiency is the ratio of output power to input power


Efficiency = Pout/Pinefficiency

EXAMPLE: Conversion of coal into electricity is through the following process. Coal burns to heat up water to steam. Steam turns a turbine. The turbine turns a generator which produces electricity. Suppose the useable electricity from such a power plant is 125 MW, while the chemical energy of the coal is 690 MW. Find the efficiency of the plant.

SOLUTION:

Efficiency = Pout/Pin

= 125 MW / 690 MW = 0.18 or 18%.

2.3.11 Solve problems involving momentum, work, energy and power.



Solve problems involving momentum, work, energy and power.

EXAMPLE: A 25-kg object resting on a frictionless incline is released, as shown. What is its speed at the bottom?

SOLUTION:

We solved this one long ago using Newton’s second law. It was difficult!

We will now use energy to solve it. ∆EK + ∆EP = 0

(1/2)mv2 - (1/2)mu2 + mg∆h = 0

(1/2)(25)v2 - (1/2)(25)02 + (25)(10)(-6) = 0

12.5v2 = 1500

v = 11 m s-1.

30°

6.0 m

u = 0

v = ?∆h



PRACTICE: A 3.0-kg ball thrown at a wall at 9.0 m s-1 (+x-dir). It rebounds at 8.0 m s-1 (–x-dir). Find its change in momentum and change in kinetic energy. Is this an elastic collision?p0 = mu = (3)(+9) = 27 kg m s-1.pf = mv = (3)(-8) = -24 kg m s-1.Thus ∆p = pf – p0 = -24 – 27 = -51 kg m s-1.

EK,0 = (1/2)mu2 = (1/2)(3)92 = 121.5 J.EK,f = (1/2)mv2 = (1/2)(3)82 = 96 J.Thus ∆EK = EK,f – EK,0 = 96 – 121.5 = -26 J.

This is not an elastic collision since ∆EK ≠ 0.



EXAMPLE: Suppose a .020-kg bullet traveling horizontally at 300. m/s strikes a 4.0-kg block of wood resting on a wood floor. How fast is the block/bullet combo moving immediately after collision?

SOLUTION:

If we consider the bullet-block combo as our system, there are no external forces in the x-direction at collision. Thus pf = p0 so that

mvf + MVf = mvi + MVi

.02v + 4v = (.02)(300) + 4(0) 4.02v = 6 v = 1.5 m/s

the bullet and the block move at the same speed after collision



EXAMPLE: Suppose a .020-kg bullet traveling horizontally at 300. m/s strikes a 4.0-kg block of wood resting on a wood floor. The block/bullet combo slides 6 m before coming to a stop. Find the friction between the block and the floor.

SOLUTION:

Here we use the work-kinetic energy theorem: ∆EK = W

(1/2)mv2 - (1/2)mu2 = fs cos (1/2)(4.02)(0)2 - (1/2)(4.02)(1.5)2 = f(6) cos 180° -4.5225 = -6f f = -4.5225/-6 f = 0.75 N.

s

f



EXAMPLE: Suppose a .020-kg bullet traveling horizontally at 300. m/s strikes a 4.0-kg block of wood resting on a wood floor. If the bullet penetrates .06 m of the block, find the average force F acting on it during its collision.

SOLUTION:

We again use the work-kinetic energy theorem on only the bullet: ∆EK = W

(1/2)mv2 - (1/2)mu2 = Fs cos (1/2)(.02)(1.5)2 - (1/2)(.02)(300)2 = -F(.06) -900 = - .06F

F = 15000 n.

sF

2.3.1outline what is meant by work. 2.3.2determine the work done by a non-constant force by...

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work w

units of work

principles of work

newton force f

force f times

displacement s

constant force

force acts