WorkWorkThe work done on an object by a constant

forceis the product of the force that is parallel to thedisplacement (θ = 0°) times the displacement.

Work is computed by

W = F × Δx × cosθ

where F is measured in N, Δx is thedisplacement measured in m and θ is the

anglebetween the force and the displacement.

The unit of work is

1N•1m = 1J

where J is the work done measured in joules.

Do not confuse a N•m which is also used to

measure torque.

Work is a scalar quantity meaning it has only direction.

When is there no work done in a “physics sense”?

When Δx = 0, W = 0 J in all cases.

No matter how long you hold a heavy package, if your displacement is zero,

so is the work done.

•

F

Fw

You are applying a force butΔx = 0 which makes W = 0.

What happens when you carry this heavyobject a displacement of 2.0 m?

You are exerting a force, you do have a displacement, but the angle between the applied force and the displacement is 90°, therefore the W = 0.

•

F

Fw

Δx

What about an object undergoing uniform horizontal circular motion?

You are exerting a force, you do have a displacement, but the angle is 90°, and the cos 90° = 0, therefore the W = 0.

Fc

Δx

Work ProblemsWork Problems

How much work would you do if you pushed a

140 kg crate 12.7 m across a floor at a constant

velocity if µ = 0.60?

m = 140 kg Δx = 12.7 m µ = 0.60

Ff F

Fw

FN Fy = 0

FN = Fw

Fx = 0

F = Ff

µ = Ff/FN

Ff = µFN = µFw = µmg

Ff = 0.60 × 140 kg × 9.80 m/s2 = 820 N

W = F × Δx × cosθ

W = 820 N × 12.7 m × 1 = 1.0 x 104 J

A 60. kg crate is pulled 50. m along a horizontal

surface by a constant force of 110. N which acts

at an angle of 40° with the horizontal. The

frictional force is 50. N.

(a) What is the work done by each force acting

on the crate.FHFf

FV

FN F

Fw

θ

m = 60. kg F = 110. N

Δx = 50. m θ = 40°

Ff = 50. N

W(Fw) = Fw × Δx cos 90° = 0 J

W(Fv) = FV × Δx cos 90° = 0 J

W(FN) = FN × Δx cos 90° = 0 J

W(FH) = FH × Δx cos 40°

W(FH) = 110. N × 50. m × cos 40° = 4.2 x 103 J

W(Ff) = Ff × Δx cos 180°

W(Ff) = 50. N × 50. m × cos 180° = -2.5 x 103 J

(b) What is the net work done on the crate?

WV = 0 J in a vertical direction.

WT = FH + Ff + WV

WT = 4.2 x 103 J – 2.5 x 103 J + 0

WT = 1.7 x 103 J

Work-Energy PrincipleWork-Energy Principle

Energy is the ability to do work.

Kinetic energy (translational energy) is theenergy resulting from the motion of an objectand is given by

KE = ½mv2 = 1 kg•1 m2/s2 = N•m = J

This is no coincidence that work and energy are measured in the same units.

The work-energy principle states that the network done on or by an object is equal to itschange in kinetic energy.

W = ΔKE = ½mv2

When ΔKE > 0, work is done on the object and its kinetic energy increases.

When ΔKE < 0, work is done by the object and its kinetic energy decreases.

Gravitational potential energy is the storedenergy an object has because of its positionrelative to the earth.

The work-energy principle states that the network done on or by an object is equal to itschange in potential energy.

W = F × Δx × cosθ

W = ΔPEg = mgΔh = kg•m/s2•m = J

Δh is the height measured from a convenientreference point.

The reference point can be anywhere because you are only interested in the change in energy, not the actual energy.

When ΔPEg > 0, work is done on the object and its potential energy increases.

When ΔPEg < 0, work is done by the object and its potential energy

decreases.

Work-Energy Principle ProblemWork-Energy Principle Problem

How much work is required to accelerate a1250 kg car from 30. m/s to 40. m/s?

m = 1250 kg vi = 30. m/s vf = 40. m/s

W = ΔKE = ½mvf2 - ½mvi

2

W = ½ × 1250 kg × (40. m/s)2 – ½ × 1250 kg × (30. m/s)2

W = 4.4 x 105 J

Because W > 0, work is done on the carincreasing its kinetic energy.

More Work-Energy ProblemsMore Work-Energy Problems

A 75 kg crate is pushed up a rough inclinedplane by a force of 470 N which is parallel to theplane. The plane is 20. m long and makes anangle of 25° with the ground.

(a) How much work is done by pushing the crate from the bottom to the top of the plane?

m = 75 kg FT = 470 N l = 20. m

θ = 25°

.

•

θ

FN

Fp

FT

Fw

FN

θ

W = F × Δx × cosθ = 470 N × 20. m × 1 = 9.4 x 103 J

Ff

(b) What is the change in potential energy of the

crate?

ΔPEg = mgh = 75 kg × 9.80 m/s2 × 20. m ×

sin 30°

ΔPEg = 7.4 x 103 J

(c) What is the work done against friction?

Fp = Fwsinθ

Fp = 75 kg × 9.80 m/s2 × .500 = 370 N

FT = Fp + Ff

Ff = 470 N – 370 N = 1.0 x 102 N

W = Ff × Δx × cosθ

W = 1.0 x 102 N × 20. m × -1 = -2.0 x 103 J

It is important to realize that something is remaining constant in the problem.

That something is the sum of the ΔPEg

(gravitational potential energy) and Wf

(work done against friction) equals the work done in pushing the crate the entire length of the plane.

A 70. g arrow is fired from a bow whose string

exerts a force of 85 N on the arrow over adistance of 90. cm. What is the speed of

thearrow as it leaves the bow?

m = 70. g Δx = 90. cm F = 85 N

W = ΔKE

F × Δx × cosθ = ½mv2

F × Δx × cosθ = ½mv2

v = (2 × F × Δx × cosθ/m)½

v = ((2 × 85 N × 90. cm × 1 m/100 cm × 1)/

70. g × 1 kg/103 g))½

v = 47 m/s

A 65 kg boy scout starts at an elevation of1200. m and climbs to an elevation of 2700. m.

(a) What is the boy scout’s change in potential energy?

m = 65 kg g = 9.80 m/s2

hi = 1200. m hf = 2700. m

ΔPEg = mgΔh = 65 kg × 9.80 m/s2 × (2700. m – 1200. m) = 9.6 x

105 J

(b) What is the minimum work?

W = ΔPEg = 9.6 x 105 J

(c) Is it possible for the work required to be

greater than the minimum amount of work?

Yes, if friction was present, WT = Wf + ΔPEg

Conservation of EnergyConservation of Energy

Gravity is considered to be a conservativeforce.

When computing ΔPEg, the value does not depend on the path but only on the initial and final positions.

Sometimes gravity is referred to as a state function because it is independent of path.

Friction is considered to be a non-conservative

force or a dissipative force.

Because friction is dependent on path, it is sometimes referred to as a path function.

Whenever friction is present, there will always be a certain amount of mechanical energy that gets “wasted” or turned into heat.

The amount of energy “wasted” due to friction does depend on the path

taken.

The conservation of mechanical energy holds

true only if there are conservative forces acting

on a system.

ΔE = 0

ΔKE + ΔPE = 0

A Pendulum ProblemA Pendulum Problem

A 0.57 kg pendulum bob at the end of a 0.90 m

cord is displaced 0.10 m and then released.Using only energy considerations, determinethe maximum speed of the bob. The center

ofgravity was raised 0.0050 m.

m = 0.57 kg l = 0.90 m Δx = 0.10 m

Δh = 0.0050 m g = 9.80 m/s2

.

Δh

ΔE = 0

(KE + PEg)amp = (KE + PEg)eq pos

0 0

.

(PEg)amp = (KE)eq pos

mgΔh = ½mv2

v = 2gΔh

v = (2 × 9.80 m/s2 × 0.0050 m)½

v = 0.31 m/s

PowerPower

Power is the rate at which work is done or the

rate at which energy is transformed into another

form of mechanical energy.

where P is the power measured in watts, W, W

is the work in joules, J, and Δt is the time inseconds, s.

P =W

Δt=

J

s= W

Power is sometimes expressed in horsepower,

1 hp = 746 W.

Power can also be expressed as

P =W

Δt=

F × Δx × cosθ

Δt=F × v × cosθ

A Power ProblemA Power Problem

A 60. kg boy runs up a flight of stairs in 3.7 s.

The vertical height of the stairs is 4.2 m.

(a) What is the boy’s output power both in watts

and horsepower?

m = 60. kg Δt = 3.7 s Δh = 4.2 mP =

WΔt

=ΔPEgΔt

=Δt

mgΔh

.

P =60. kg × 9.80 m/s2 × 4.2 m

3.7 s

P = 670 W ×1 hp

746 W=0.90 hp

(b) How much energy is involved?

P =E

Δt

E = 670 W × 3.7 s = 2.5 x 103 J

Elastic Potential EnergyElastic Potential Energy

To stretch or compress a spring an amount sfrom its equilibrium position requires a force

F.

Hooke’s Law is given by

Fr = -ks

where Fr is the restoring force in N, k is thespring constant in N/m, and s is thedisplacement of the spring from the

equilibriumposition.

The negative sign indicates the restoring force of the spring.

The restoring force of the spring is always trying to restore the spring to its equilibrium position.

The greater the value of k, the stiffer the spring and the more difficult it is to stretch or compress it.

Because Fr α s, the more a spring iscompressed or stretched, the greater therestoring force Fr.

W = ½•F•s•cosθ

The cos θ always equals 1 and Fr = ks

which gives

W = ΔPEe = ½ks2

Applying the Conservation of MechanicalEnergy to a spring gives

ΔE = 0

(PEe + KE)amp = (PEe + KE)eq pos

0 0

PEe = KE

½ks2 = ½mv2

Spring ProblemSpring Problem

A spring stretches 0.150 m when a 0.300 kg

mass is suspended from it. The spring isstretched an additional 0.100 m from thisequilibrium position and released.

0.150 m0.100 m

0.300 kg

0.300 kg

(a) Determine the spring constant.

m = 0.300 kg s = 0.150 m Δx = 0.100 m

Fr = ks

k = Fr

s=

Fw

s

mg

s=

k = 0.300 kg × 9.80 m/s2

0.150 m=19.6 N/m

(b) What is the amplitude?

The amplitude is 0.100 m which is the maximum displacement from the

equilibrium position.

(c) Determine the maximum velocity.

(PEe + KE)eq pos = (PEe + KE)amp

0 0

½mv2 = ½kΔx2

v = (kΔx2/m)½

v = (19.6 N/m × (0.100 m)2/0.300 kg)½

v = 0.808 m/s in the direction of either amplitude

(d) Determine the velocity when the mass is 0.050 m from the equilibrium position.

(PEe + KE)amp = (PEe + KE)disp

½kΔx2 = ½ks2 + ½mv2

19.6 N/m × (0.100 m)2 = 19.6 N/m × (0.050 m)2 + 0.300

kg × v2

0

v = 0.700 m/s in the direction of either amplitude

(e) Fnet = ma

a = Fnet/m = Fr/m = kΔx/m

a =19.6 N/m × 0.100 m

0.300 kg=6.53 m/s2

Wrap Up QuestionsWrap Up Questions

A mass that is attached to the free end of aspring undergoes simple harmonic motion.

Does the total energy change if the mass isdoubled and the amplitude remains the same?

The total energy is given by PEe = ½kxmax2,

therefore changing the mass has no effect onthe total energy.

Does the kinetic energy depend on mass?

Yes, because kinetic energy, KE, is given byKE = ½mv2.

If the net work done on an object is equal tozero, what is true about the object’s

velocity?

The velocity remains the same becauseW = ΔKE = 0 meaning that vf = vi.

Why is the work done by a frictional force, Ff, is

always negative?

The frictional force always opposes thedisplacement of the object making the

anglebetween the force and displacement

180°, andthe cos 180° = -1.

Can the average power ever equal theinstantaneous power?

Yes, if the power remains constant throughout a

time interval.

Using the work-energy principle, show why

friction always reduces the kinetic object of an

object.

Because the angle between the displacement

and the frictional force is always 180°,

W = ΔKE = ½mvf2 - ½mvi

2 < 0 because vf < vi.