Chapter 6: Work, energy , and power 6-1 Work 6-2 Kinetic energy 6-3 Potential energy and conservative forces 6-4 Di

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Chapter 6: Work, energy , and power 6-1 Work 6-2 Kinetic energy 6-3 Potential energy and conservative forces 6-4 Dissipative forces 6-9 power. 6-1 Work. W = F. Δ x. units: N.m or joule, j. W = F Δ x cos q

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## Chapter 6: Work, energy , and power 6-1 Work 6-2 Kinetic energy 6-3 Potential energy and conservative forces 6-4 Di

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Chapter 6: Work, energy , and power

6-1 Work

6-2 Kinetic energy

6-3 Potential energy and conservative forces

6-4 Dissipative forces

6-9 power

Norah Ali Al Moneef king Saud university

6-1 Work

W = F.Δx

units: N.m or joule, j

• W = F Δx cos q
• Work = the product of force and displacement, times the cosine of the angle between the force and the direction of the displacement.
• How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m?
• W = FΔx cos q
• W = 15 N (12 m) (cos 20)

15 N

20o

= 169J

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Sign of work:

+vework: Force acts in the same direction as the displacement.

 object gains energy

ve work: Force acts in the opposite direction as the displacement.

 object loses energy

When force F is at right angles to displacement s (F ⊥s):

 cos 90 = 0

• No work is done if:
• F = 0 or
• Δx = 0 or
• q= 90o

 F Δxcos = 0

 No work is done on the load

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example

• A 1500 kg car moves down the freeway at 30 m/s
• K.E. = ½(1500 kg)(30 m/s)2= 675,000 kg·m2/s2 = 675 kJ
• A 2 kg fish jumps out of the water with a speed of 1 m/s
• K.E. = ½(2 kg)(1 m/s)2= 1 kg·m2/s2 = 1 J
• Calculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cm.
• work = force x distance
• the child must exert an upward force equal to its weight
• the distance moved upwards equals (12 x 20cm) = 2.4m
• work = 300 N x 2.4 m= 720 J

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6-2 Kinetic Energy

Same units as work

Remember the Eq. of motion

Multiply both sides by m,

W = F Dx

W = DK

Work-Energy Theorem

• When work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system.
• The speed of the system increases if the work done on it is positive
• The speed of the system decreases if the net work is negative

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Kinetic Energy – the energy of motion
• KE = ½ m v 2
• units: kg (m/s) 2 = (kgm/s2) m
• = Nm = joule
• Work Energy Theorem: W = D KE
• Work = the change in kinetic energy of a system.

Note: v = speed NOT velocity.

The direction of motion has no relevance to kinetic energy.

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Factors Affecting Kinetic Energy
• The kinetic energy of an object depends on both its mass and its velocity.
• Kinetic energy increases as mass increases.
• For example, think about rolling a bowling ball and a golf ball down a bowling lane at the same velocity, as shown in Figure 2.
• The bowling ball has more mass than the golf ball.
• Therefore, you use more energy to roll the bowling ball than to roll the golf ball.
• The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball.

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Kinetic energy also increases when velocity increases.
• For example, suppose you have two identical bowling balls and you roll one ball so it moves at a greater velocity than the other.
• You must throw the faster ball harder to give it the greater velocity.
• In other words, you transfer more energy to it.
• Therefore, the faster ball has more kinetic energy.

Figure 2Kinetic Energy Kinetic energy increases as mass and velocity increase. Predicting In each example, which object will transfer more energy to the pins? Why?

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Calculating Kinetic Energy
• There is a mathematical relationship between kinetic energy, mass, and velocity.

Example : A 6 kg bowling ball moves at 4 m/s.a) How much kinetic energy does the bowling ball have?b) How fast must a 2.5 kg tennis ball move in order to have the same kinetic energy as the bowling ball?

K = ½ mb vb²

K = ½ (6 kg) (4 m/s)²

KE = 48 J

KE = ½ mt vt²

v = √2 K/m = 6.20 m/s

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Do changes in velocity and mass have the same effect on kinetic energy?
• No—changing the velocity of an object will have a greater effect on its kinetic energy than changing its mass.
• This is because velocity is squared in the kinetic energy equation.
• For instance, doubling the mass of an object will double its kinetic energy.
• But doubling its velocity will quadruple its kinetic energy.

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example

Calculate the speed of a car of mass 1200kg if its kinetic energy is 15 000J

K = ½ m v2

15 000J = ½ x 1200kg x v2

15 000 = 600 x v2

15 000 ÷ 600 = v2

25 = v2

v = 25

speed = 5.0 ms-1

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Calculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN.

k.E. of car = ½ m v2

= ½ x 900kg x (20ms-1)2

= ½ x 900 x 400

= 450 x 400

k = 180 000 J

The work done by the brakes will be equal to this kinetic energy.

W = F Δx

180 000 J = 3 kN x Δx

180 000 = 3000 x Δx

Δx = 180 000 / 3000

braking distance = 60 m

example

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6-3 Potential energy and conservative forces
• An object does not have to be moving to have energy.
• Some objects have stored energy as a result of their positions or shapes.
• When you lift a book up to your desk from the floor or compress a spring to wind a toy, you transfer energy to it.
• The energy you transfer is stored, or held in readiness.
• It might be used later when the book falls to the floor
• Stored energy that results from the position or shape of an object is called potential energy.
• This type of energy has the potential to do work.

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Gravitational Potential Energy
• Potential energy related to an object’s height is called gravitational potential energy.
• The gravitational potential energy of an object is equal to the work done to lift it.
• Remember that Work = Force × Distance.
• The force you use to lift the object is equal to its weight.
• The distance you move the object is its height.
• You can calculate an object’s gravitational potential energy using this formula:
• Gravitational Potential Energy = Weight x Height

change in Gravitational potential energy .

= mass x gravitational field strength x change in height

ΔU = m g Δh

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Gravitational potential energy:

• Potential energy only depends on y (height) and not on x (lateral distance)
• MUST pick a point where potential energy is considered zero!

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example

Calculate the change in Gravitational potential energy (g.p.e). when a mass of 200 g is lifted upwards by 30 cm.

(g = 9.8 Nkg-1)

ΔU = m g Δh

= 200 g x 9.8 Nkg-1 x 30 cm

= 0.200 kg x 9.8 Nkg-1 x 0.30 m

change in g.p.e. = 0.59 J

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example

Q: Calculate the gravitational potential energy in the following systems

a. a car with a mass of 1200 kg at the top of a 42 m high hill

(1200 kg)( 9.8m/s/s)(42 m) = 4.9 x 105

b. a 65 kg climber on top of Mt. Everest (8800 m high)

(65 kg) (9.8m/s/s) (8800 m) = 5.6 x 106 J

c. a 0.52 kg bird flying at an altitude of 550 m

(.52 kg) (9.8m/s/s)(550) = 2.8 x 103 J

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Conservation of Energy
• Law of conservation of energy

- energy cannot be created or destroyed

closed system:

- all energy remains in the system

- nothing can enter or leave

open system:

- energy present at the beginning of the system may not be at present at the end

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Conservation of Energy
• When we say that something is conserved it means that it remains constant, it doesn’t mean that the quantity can not change form during that time, but it will always have the same amount.
• Conservation of Mechanical Energy:

MEi = MEf

• initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravity:

Ko + Uo = K + U

½ mvo² + mgho = ½ mv² + mgh

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example

• At a construction site a 1.50 kg brick is dropped from rest and hit the ground at a speed of 26.0 m/s. Assuming air resistance can be ignored, calculate the gravitational potential energy of the brick before it was dropped?
• K+U = Ko+Uo
• ½ mv2+mgh=½ mv2o+ mgho
• ½ mv2+0 = =o+ mgho
• U0 = mgho = ½ (1.50 kg) (26.0m/s)2 = 507 J

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A child of mass 40 kg climbs up a wall of height 2.0 m and then steps off. Assuming no significant air resistance calculate the maximum:

(a) gravitational potential energy (gpe) of the child

(b) speed of the child

g = 9.8 Nkg-1

(a)maximum gravitational potential energy ( max gpe) occurs when the child is on the wall

Δ U = mgΔh

= 40 x 9.8 x 2.0

max gpe = 784 J

(b) max speed occurs when the child reaches the ground

½ m v2= m g Δh

½ m v2= 784 J

v2= (2 x 784) / 40

v2= 39.2

v = 39.2

max speed = 6.3 ms-1

example

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example

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distance.

m = 10.0 kg vi = 0

Fp = 250.0 N Fk = 50.0 N

Δx = 10.0 m

FN

A. How much work is done by each force on the cart?

Fk

Fp

Wg = 0 WN = 0

Fg

Wp = FpΔx cos = (250.0 N)(10.0 m)cos 0

Wp = 2500 J

Wk = FkΔx cos= (50.0 N)(10.0 m)cos 180= -500 J

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B. How much kinetic energy has the cart gained?

Wnet = ∆KE

Wp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0

KEf = 2000J

C. What is the carts final speed?

KE = 1/2 m v2

v = √((2KE)/(m))

v = √((2(2000 J))/(10.0 kg)) v = 20 m/s

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6-4 Dissipative forces

Nonconservative Forces
• A nonconservative force does not satisfy the conditions of conservative forces
• Nonconservative forces acting in a system cause a change in the mechanical energy of the system
• The work done against friction is greater along the brown path than along the blue path
• Because the work done depends on the path, friction is a nonconservative force

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example

On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does it travel if the coefficient of kinetic friction between the sled and the ice is .10?

m = 10.0 kg

vi = 2.2 m/s

vf = 0 m/s

µk = .10

W = ∆KE

FΔx= Kf - Ki

µk = Fk / FN

Fk = µk (mg)

µk (-mg) Δx = 1/2 m vf2 - 1/2 m vi2

µk (-mg) Δx = - 1/2 m vi2

Δx = (- 1/2 m vi2) / µk (-mg)

Δx = (- 1/2 (10.0 kg) (2.2 m/s)2) / (.10 (-10.0 kg) (9.8 m/s2))

Δx = 2.47 m

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Example

A 20-g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s.

W =Δ K

F Δ x cosθ = K – K0

F Δ x cosθ= ½ mv² - ½ mvo²

F (0.06 m) cos 1800 = 0 - 0 .5 (0.02 kg)(80 m/s)2

F (0.06 m)(-1) = -64 J F = 1067 N

Work to stop bullet = change in K.E. for bullet

W = ½ mv² - ½ mvo²

= 0

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example

A bus slams on brakes to avoid an accident. The tread marks of the tires are 80 m long. If μk = 0.7, what was the speed before applying brakes?

Work = F Δx (cos θ)

f = μk.N = μk mg

Work = - μk mg Δx

ΔK = 1/2 mv2 - ½ mvo2

W = ΔK

vo = (2μk g Δ x) ½

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6-9 power
• Power is the rate at which work is done.
• Power is the rate of energy transfer by any method.
• The SI unit of power is the watt, W
• 1 watt = 1 Joule/s
• 1000 watts = 1 kW

Work= force · distance

Power = force·distance/time

Power = force·velocity

P = F·v

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example

Calculate the power of a car engine that exerts a force of 40 kN over a distance of 20 m for 10 seconds.

W = FΔx

= 40 kN x 20 m

= 40 000 x 20 m

= 800 000 J

P = ΔW / Δt

= 800 000 J / 10 s

power = 80 000 W

• In 1935, R. Goddard, the America rocket pioneer, launched several A-series sounding rockets. Given that the engine had a constant thrust of 890 N, how much power did it transfer to the rocket while traveling at its maximum speed of 113.9 m/s.
• P = Fv = 890N x 313.9 m/s = 279 kW.

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example
• Together, two students exert a force of 825 N in pushing a car a distance of 35 m.
• A. How much work do they do on the car?
• W = F Δx = 825N (35m) =
• B. If the force was doubled, how much work would they do pushing the car the same distance?
• W = 2F Δx = 2(825N)(35m) =

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example

• A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point.
• How much work does the climber do on the backpack?
• W = Fd = m g Δx= 7.5kg(9.8m/s2)(8.2m)
• b.How much power does the climber expend in this effort?
• P = W = 603 J = 0.34 watt
• t 1800 s

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Example : A 200 kg curtain needs to be raised 8m. in as close to 5 s as possible. You need to decide among three motors to buy for this, each motor cost a little more the bigger the power rating. The power rating for the three motors are listed as 1.0 kw, 3.5 kw and 5.5 kw. Which motor is the best for the job?

• Given: m = 200 kg Δx = 8 m ∆t = 5s
• Unknown: Power and work

W = F·Δx

W = m·g·h

W = (200 kg)·(9.81 m/s²)·(8 m) = 15,696 Joules

P = W/∆t

P = 15,696 J / 5 s= 3,139 watts

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Calculate the power of a car that maintains a constant speed of 30 ms-1 against air resistance forces of 2 kN

As the car is travelling at a constant speed the car’s engine must be exerting a force equal to the opposing air resistance forces.

P = F v

= 2 kN x 30 ms-1

= 2 000 N x 30 ms-1

power = 60 kW

example

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A 100-kg cheetah moves from rest to 30 m/s in 4 s. What is the power?

the work is equal to the change in kinetic energy:

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Summary

• If the force is constant and parallel to the displacement, work is force times distance
• If the force is not parallel to the displacement,
• The total work is the work done by the net force:
• SI unit of work: the joule, J
• Total work is equal to the change in kinetic energy:

where

• Power is the rate at which work is done:
• SI unit of power: the watt, W

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