Chapter 6

Chapter 6

Work and Energy

Younes Sina

Work (W): Work is defined as the product of parallel force and distance. W = F ll x .F ll denotes the component of force F that is parallel to displacement , x .The SI unit for work is N.m called " Joule", and the American unit is " lb-ft "If the direction of F is parallel to that of x, the work done is simplyW = F x.

Example :Find the work done by force F = 25N in pushing the block a distance of 14m .

25 N force appliedBlock moved by 14 m

25 N

14 m

W=F . xW=(25N)(14 m)W=350 Joules

Solution

If an object placed on a horizontal surface is pushed or pulled by force F as shown below, the component of F that is parallel to x does useful work .The component perpendicular to x does no work.

Example : A block is pulled a distance of x = 24m from A to B as shown via force F = 45N that makes a 30 .0 o angle with the horizontal surface . Calculate the work done by F.

x = 24m

30 .0 o

Solution: For θ = 30.0o, and F = 45NF ll = 45N cos (30 .0 ) = 39NW = F ll xW = (39N)(24m) = 940 NmW = 940 J

Example : In the following figure, find (a) the magnitude of F such that the block slides at constant speed to the right .(b) Find the work done by this force if the displacement is 38 m .

Example: In previous example, find the work done by the frictional force, F k , within the same distance.

Solution: F k acts opposite to x; therefore, it does negativework.

W friction = F k x

W = ( -45N )( 38m ) = -1700 J

Energy: Energy is defined as the ability to do work. Energy and work are expressed in same units. Typical units are: J, cal, kcal, and eV.

Types of Energy:Energy exists in different forms such as mechanical, electric, nuclear, light, chemical, etc... .In this chapter, mechanical energy will be discussed only.Heat energy is also a will be discussed in Chapter 12 .Mechanical Energy:Kinetic energy, Gravitational potential energy, and elastic (spring) potential energy are forms of mechanical energy and will be discussed under this topic.

I ) Kinetic Energy (K.E.):Kinetic energy is the energy that a mass (an object) has because of its motion. As long as an object moves or has some speed, it has kinetic energy.K.E. is proportional to mass ( M) and proportional to the square of velocity ( v 2 ) .K.E. = (1/2) Mv 2

II ) Gravitational Potential Energy (P.E.):This energy is the type that an object has due to its elevation with respect to a reference level. This energy is significant when objects are in the vicinity of planets, stars and other heavenly masses .Gravitational P.E. is proportional to the acceleration of gravity ( g) ofthe planet or star, the mass of the object, M, and its elevation ( h) from a reference level.P.E. = Mgh

Example :Calculate the K .E . of a 2000-kg car that is moving(a) at 10.0m/s(b) 20.0 m/s(c) 30.0 m/s

Solution: (3 sig. fig.)

(a) (K.E.)1 = (1/2)(2000.kg)( 10 .0m/s) 2 = 100,000 (kg m/s 2)m = 100,000 J(b) (K.E.)2 = (1/2)(2000.kg)( 20 .0m/s) 2 = 400,000 (kg m/s 2)m = 400,000 J(c) (K.E.)3 = (1/2)(2000.kg)( 30 .0m/s) 2 = 900,000 (kg m/s 2)m = 900,000 J

Example:A rock of mass 6.1 kg is raised to the top of a building 13 m high.Calculate its gravitational P .E . with respect to the ground.

Solution:P.E. = MghP.E. = (6 .1 kg)(9 .8 m/s 2)(13 m)P.E. = 780 Nm = 780J

Example :In the following figure, a 95-kg crate is pushed up on an incline that is practically frictionless. The incline is 8 .0m long and makes a 31 o

angle with the horizontal floor. Calculate the P .E . Of the crate with respect to the floor when it reaches the top of the incline.

Example : A swimming pool is on the top of a hill at an average elevation of 150mfrom a certain ground level . The swimming pool has dimensions:15 m X 25 m X 2 .4 m and is full. The mass density of water is 1.000 ton/m 3. Determine(a) the P .E . Of the water in the pool with respect to that level .If the pool is allowed to empty to the ground level,(b) how much energy is at most available for use?(c) how much electric energy will become available? If a generator uses this energy for electricity production, and the overall efficiency is 62%

Solution: (a) The pool's volume is V = w l hV = 900m 3

r = M / VM = r VM = (1000 kg/m 3)(900 m3) = 900,000 kgP.E. = MghP.E. = ( 900,000 kg )( 9 .8 m/s 2 )( 150m ) = 1.3 × 10 9 J(b) 1.3 × 10 9 J(c) 0.62 (1.3 × 10 9 J) = 8.2 × 10 8 J

100 W

2278 hr

Example : A 750-kg car is traveling at a velocity of 72 km/h eastwardand on a level road . Determine (a) its initial kinetic energy (K .E .) i

If this car is slowed down to a speed of 36 km/h, calculate (b) its final (K.E.) f .(c) How much is the change in its kinetic energy and where does thisenergy go ?

Vi=72 km/h Vf=36 km/h

Solution:(a) (K.E.) i = (1/2)Mv i

2

(K.E.) i = (1/2)( 750 kg)( 20 . m/s)2 = 150,000 J(b) (K.E.) f = (1/2)Mv f

2

(K.E.) f = (1/2)( 750 kg)( 10 . m/s)2 = 38,000 J(c) Δ(K.E.) = (K.E.) f - (K.E.) i

Δ(K.E.) = 38,000 J - 150,000 J = - 112,000 J

This energy is consumed by force of friction (brakes force, for example) that acts opposite to the direction of motion . In fact, it is the work done by force of friction. Most of this energy converts to heat and warms up the brakes. Friction does negative work .

Example: If the work done by force of friction is -112,000J .Use the work formula to calculate the force applied by friction if brakes were used within a distance of 56m .

Wf=-112,000J

56m

Solution:W = F k x- 112,000 J = F k (56m)F k = - 2000N

Work-Kinetic Energy Theorem:

The work done by the net force acting on a mass is equal to the change in the kinetic energy of that mass.

(ΣF ) x = Δ(K.E.)

(ΣF ) x = (K.E.) f - (K.E.) i

Example : A 900-kg car traveling at 15m/s changes its speed to 25m/s in a distance of 50 m due to a net force . Calculate(a) the net force(b) the engine force if the frictional forces add up to 1400N .

V=15m/s V=25m/s

X=50 m

M= 900 kg

Solution:(a) Work-K.E. theorem :(Σ F) ∙ x = (K.E.) f - (K.E.) i

Σ F (50 m) = (1/2)(900)(25) 2 - (1/2)(900)(15) 2

Σ F = 3600 N(b) Σ F = F e - Ff

3600N = Fe - 1400NF e = 5000 N

Example : A boy pulls a 45.0-kg sled (including his friend in it)from rest for a distance of 12.0 m with a horizontal and constant force of 165.0 N. The frictional force between the sled and snow is 55.0 N. Calculate (a) the work done by the boy on the sled(b) the work done by the frictional force on the sled(c) the work done by the net force on the sled(d) the speed of the sled at the end of the 12.0 m distance .

12.0m

45.0 kg

165.0 N

Wf=55.0 N

v=0 m/s v=? m/s

Solution:

(a) Wboy = F ll xW = (165 N)(12 .0 m) = 1980 J(b) Wfriction = F k xW = (55 .0 N)(12.0 m) = 660.0 J(Work of friction is negative)(c) Wnet = Fnet x(Σ F) x = (165N - 55.0 N)(12.0 m) = 1320 J(d) (Σ F) x = (K.E.) f - (K.E.) i

1320 J = (1/2)(45 .0kg) Vf2 - (1/2)(45 .0kg) (0)2

Vf = + 7.66 m/s

The Law of Conservation of Energy

This law states that:

"Energy is conserved.

It is neither created nor destroyed.

It converts from one form to another".

Conservation of Mechanical Energy

Example : In the following figure, neglecting friction, find the speed Of the 750-kg car at the bottom of the hill . Suppose the car is put inneutral and starts from rest from the top of the hill .

V=0, M=750 kg

45 m

Conservation of mechanical energy:Total Energy at A = Total Energy at B(K .E .)A + (P .E .)A = (K .E .)B + (P .E .)B

(1/2)MVA2 + M g hA = ( 1/2)MVB

2 + M g hB

VA = 0 and hB = 0M g hA = ( 1/2)MVB

2

g hA = (1/2)VB2

2 g hA = VB2

VB = 30 m/s

A

BhA

hB = 0

VB

Example :In the following figure, if 120,000 J of energy is consumed byfrictional forces, find the speed of the 750-kg car at the bottom of the hill . Suppose the car is put in neutral and starts from rest from the top of the hill .

frictional Energy=120,000 J

In this case, part of the available energy will be wasted by friction

Conservation of mechanical energy:Total Energy at A - W friction= Total Energy at B(K .E .)A + (P .E .)A - Wfriction = (K .E .)B + (P .E .)B

(1/2)M A2 + M g hA -120,000J = ( 1/2)MV B

2 + M g hB

V A = 0 and h B = 0M ghA -120,000J = ( 1/2)MV B

2

(750)(9 .8)(45) J - 120,000 J = 0 .5(750 kg)VB2

562(m/s) 2 = VB2

VB = 24 m/s

V=0, M=750 kg

Example: In the figure shown, find(a) the work done by the frictional force on the car as it coasts down the hill in neutral(b) the energy loss due to friction(c) the speed of the 750-kg car (VB) at the bottom of the hill .

400 m

Solution:(a) The work done by friction isW friction = F k xW friction = ( -300N)(400m) = -120,000 J(b) -120,000 J(c) Conservation of mechanical energy:Total Energy at A - Wfriction = Total Energy at B(K .E .)A + (P .E .)A - Wfriction= (K .E .)B + (P .E .)B

(1/2)MVA2 + M g h A - 120,000 J = (1/2)MVB

2 + M g hB

VA = 0 and hB = 0M ghA -120,000 J = (1/2)MVB

2

(750)(9.8)(45) J - 120,000 J = 0.5(750)VB2

[562(m/s)]2 = VB2

VB = 24 m/s

Power is defined as the work done per units of time. Mathematically:

The (SI) unit for power is :The industrial unit: hp

Power

t

WP

=

Wattcalled s

J

s

ftlbhphP

== 5501 and watts7461

Example : An electric motor is capable of delivering 7 .2 Million Joules of work in one hour and 400.0 seconds. Find the power of the motor in watts, kilowatts, and hp .

Solution:P = W/tP = (7 .2 x10 6 J )/(4000 s) = 1800 J/s = 1800 wattsP = 1.8 kilowatts (kw)P = (1800/746 ) hpP = 2.4 hp

Example:Calculate the amount of work or energy that a 4.50 hpelectric motor can deliver in 10.0 hours.

Solution:P = 4 .5hp = (4 .50)(746 watts) = 3360 watts or ( J /s )P = W / tW = PtW = (3360 J/s)(36,000s) = 1.21 x 10 8 J

Efficiency: When a device receives power from a source, it does not deliver all of it in the intended form and converts a portion of it to other undesired forms. This makes a device to be less than 100 percent efficient.

Most power plants have efficiencies of under 50 percent. A good nuclear power plant is about 45% efficient . Early coal plants were hardly 20% efficient .

Eff. = Pout /Pin

Example : A 1.25 hp electric pump with an efficiency of 92.5% is used to perform1 .40 Mega Joules of work.(a) How long does it take for the electric pump to do the job ?(b) If it was 100% efficient , how long would it take?

Solution:A 1.25 hp electric pump means it draws 1 .25 hp of electric power from the source.It generally cannot give back all of it .Make sure that you write the formulas with horizontal fraction bars .Eff. = P out / P in

0 .925 = P out / (1 .25 hp)P out = 0 .925(1 .25hp) = 1.16 hp(a) P = W / tt = W / Pt = (1 .40 x 10 6 J) / ( 1 .16 x 746 watts ) = 1620 sec(b) P = W/tt = W/Pt =(1 .40 x 10 6 J)/( 1 .25 x 746 watts ) = 1500 sec

Example: Calculate (a) the necessary power to empty a swimming pool with dimensions (12m)(25m)(2 .4m) by pumping its water to an average height of 3.5m in 2.5hrs .The mass density of water is 1000 kg/m 3.(Hint: The power you calculate is the power that the electric pump has to deliverto water, Pout) .If the efficiency of the electric pump used is 88% , calculate (b) the electric power that the electric motor pulls from the power source, Pin .

Solution:V = w. l. hV = 720m3

r = M / VM = r VM = (1000 kg/m 3)(720 m3) = 720,000 kg(a) P = W / tW = F xF = the weight of water= Mgx = heightP = (F x) / tP = ( Mg x) / tP = (720,000kg )( 9 .8 m/s 2 )( 3 .5m ) / (2 .5 X 3600s ) = 2700 wattsP = 2.7kwP = (2700 / 746) hp = 3.6 hp(b) Eff. = P out / P in

0 .88 = 2700watts / P in

P in = 2700watts / 0 .88 = 3100 watts

Another Version of Power Formula:P = W / tW = F x P = (F x) / t(x / t )= vP = F vThis means that power is equal to force times velocity (if both of force and velocity are in the same direction).

As you see the product of velocity and force is a constant if a constant power is available . When P is constant, the product F ∙v is constant . When a greater F is needed, a smaller v is attainable and vice versa .

Example :A car is using 56 hp of its total power. When it is going up a steep hill, a force of 4200 N is needed for constant velocity motion of it up that hill. When it is traveling along a level and horizontal road, a force of 1200 N is needed for its constant velocity motion. Find the velocity of the car in each case .

P = F v

56 hp

56 hp

F= 4200 N

F=1200 N

Solution:

P = (56)(746 watts) = 42000 watts

(a) P = F vv = P / Fv = 42000 watts / 4200N = 10 m/s

(b) P = F vv = P / Fv = 42000watts / 1200N = 35 m/s

problems 1 through 5 chapter 6