Energy, Work & Power

show understanding that kinetic energy, elastic potential energy, gravitational potential energy, chemical potential energy and thermal energy are examples of different forms of energy

state the principle of the conservation of energy apply the principle of the conservation of energy to new situations or to

solve related problems state that kinetic energy Ek = ½ mv2 and gravitational potential energy

Ep = mgh (for potential energy changes near the Earth’s surface) apply the relationships for kinetic energy and potential energy to new

situations or to solve related problems recall and apply the relationship work done = force x distance moved in the

direction of the force to new situations or to solve related problems recall and apply the relationship power = work done / time taken to new

situations or to solve related problems

B

A UNIT 6

Work is done only when a Work is done only when a forceforce produces motion in produces motion in the direction of the force.the direction of the force.

Who is doingWho is doingwork ?work ?

Wall does not moved. ∴ Man is not doing work!!!

Motion is not in the direction of the force. ∴ Man is not doing work in carrying the parcel!!!

Motion

F No Motion

F

Work is the Work is the product of the force on a bodyproduct of the force on a body and and thethedistancedistance movedmoved in the direction of the force.in the direction of the force.SI unit for workSI unit for work is joule.is joule.

Work = Force x Distance moved in theWork = Force x Distance moved in thedirection of the forcedirection of the force

WW = F x d= F x d(J)(J) (N) (m)(N) (m)

Definition of One Joule:Definition of One Joule:One jouleOne joule of work is done when a force of of work is done when a force of one newtonone newtonmoves an object through a distance ofmoves an object through a distance of one meter inone meter inthe direction of the force.the direction of the force.

Moment of a force:Moment of a force:MM = Force x = Force x PerpendicularPerpendicular distance distance

from pivotfrom pivot

Units: Units: NmNm = N= N mm

WorkWorkWW = Force x distance = Force x distance along directionalong direction

of forceof force

Units: Units: JJ = N= N mm

‡ Moment of force‡ Moment of force

40m

Eg. 1 An object of weight 5N is pulled a distance of 40 m by a

force of 20N and the object moves in the same direction as the force. Calculate the work done by the force.

Solution:Solution:Work Done = Work Done = F x F x ddistanceistance moved in the direction of the moved in the direction of the appliedapplied forceforce

= 20 = 20 NN x 40x 40mm

20N

5N

= 800 = 800 NmNm= 800 = 800 J J

FORCE Distance Moved in the

direction of the force

Work Done

1000N 2m

400N 1200J

50cm 4000J

More examples…

Eg. 2A box of mass 1kg is pulled 50cm along a level floor by a

horizontal force 5N. The box is then raised vertically onto a table 80cm high. What is the total work done on the box? (Take g = 10N/kg)

0.5m

5N

10N

5N

10N

0.8m

0.5m

5N

10N

5N

10NSolution:Solution:Work Done by horizontal force = ( 5Work Done by horizontal force = ( 5NN ) x (0.50) x (0.50mm))

= 2.5 = 2.5 JJ

= 10 = 10 NNWeight of box = 1Weight of box = 1kgkg x 10 x 10 N/kgN/kg

Minimum force required to raise box verticallyMinimum force required to raise box vertically = 10 = 10 NN

5N

10N

0.8m

Solution:Solution:Work Done by vertical force Work Done by vertical force

= 10 = 10 N N x 0.80 x 0.80 mm

Total work done = 2.5 Total work done = 2.5 J J + 8 + 8 JJ

= 8 = 8 JJ

= 10.5 = 10.5 JJ

a. Gravitational potential energy (GPE) is the energy a. Gravitational potential energy (GPE) is the energy a bodya bodypossesses due to its position relative to the centre of possesses due to its position relative to the centre of the earth.the earth.

b. The b. The higherhigher the object, the the object, the greater its GPEgreater its GPE..

Gravitational Potential Energy = Gravitational Potential Energy = mghmgh (Joule)(Joule)where: mwhere: m = mass (kg)= mass (kg)

hh = height of object above reference level (m)= height of object above reference level (m)gg = gravitational acceleration =10 m/s= gravitational acceleration =10 m/s22

6m

Solution:Solution:

a. Weight lifted = a. Weight lifted = mgmg= 40 = 40 kgkg x 10 x 10 N/kgN/kg= 400 = 400 NN

b. Gained in GPE = b. Gained in GPE = mghmgh= 40 = 40 kgkg x 10 x 10 N/kgN/kg x 6 x 6 mm= 2400 = 2400 Nm Nm = 2400 = 2400 J J

Eg. 1A mass of 40 kg was lifted through a vertical distance of6 metres in a time of 5 seconds. (Take g = 10N/kg) Calculatea. the weight lifted.b. the potential energy gained by the mass

Reference level

Eg. 2 What is the gravitational potential energy of a 400g apple

3 m above the ground? (Take g = 10ms-2)

Solution:Solution:

GPE of apple = GPE of apple = mghmgh

= 0.400 = 0.400 kgkg x 10 x 10 N/kg N/kg x 3 x 3 mm

= 12 = 12 NmNm= 12 = 12 JJ

3m

Eg. 3A man uses a pulley to move a box along a slope. The box weighs 400N and

moves a distance of 2 m along the slope when the man pulls on the taut rope with a force of 220N.

a. What is the work done by the man?b. What is the gain in GPE?c. c. Explain why the work done by the man is larger than the gain in Explain why the work done by the man is larger than the gain in GPE?GPE?

2m

220N

1m

2m

220N

1m

Solution:Solution:

a. Distance moved by the box along the slope when a. Distance moved by the box along the slope when the man pulls on the rope = the man pulls on the rope = 2m2m

Work Done by the man = F x d Work Done by the man = F x d = 220= 220 NN x 2 x 2 mm= 440= 440 JJ

b. State the vertical height moved by the box = b. State the vertical height moved by the box = 1m1m

Gain in GPE = Gain in GPE = mghmgh= 400 = 400 N N x 1 x 1 mm= 400 = 400 JJ

c. Explain why the work done by the man is larger than c. Explain why the work done by the man is larger than the gain in GPE? the gain in GPE?

Energy is used to overcome friction between the Energy is used to overcome friction between the block and the slope. block and the slope.

1. Kinetic energy (KE) is the energy 1. Kinetic energy (KE) is the energy a body posses due toa body posses due toits movement.its movement.

2. Any object which 2. Any object which is moving has kinetic energyis moving has kinetic energy..3. Motion may be 3. Motion may be translationaltranslational or or rotationalrotational..4. The 4. The fasterfaster the object moves, the the object moves, the greater its KEgreater its KE..

Translational Kinetic Energy = ½mvTranslational Kinetic Energy = ½mv2 2 (Joules)(Joules)

where:where: m= mass (kg)m= mass (kg)v= velocity or speed of objectv= velocity or speed of object (m/s)(m/s)

Eg. 1

A small car of mass 1000 kg travelling at 10 m/s. Calculatethe

a. kinetic energy of the car.

b. The car is brought to rest by applying brakes. What was the final kinetic energy of the car?

c. Energy is always conserved. Explain why the initial kinetic energy of the car was not equal to the final kinetic energy.

10 m/s 0 m/s

10 m/s 0 m/s

Solution:Solution:

= ½ x 1000 x (10)= ½ x 1000 x (10)22a. Initial kinetic energy of the car = ½mva. Initial kinetic energy of the car = ½mv22

= 50 000= 50 000JJb. Final kinetic energy of the car = ½mvb. Final kinetic energy of the car = ½mv22

= ½ x 1000 x (0)= ½ x 1000 x (0)22= 0= 0JJ

c. All the kinetic energy has been converted to heat energy c. All the kinetic energy has been converted to heat energy (to overcome (to overcome frictionfriction / / air resistanceair resistance))

Eg. 2

A bullet of mass 40g leaves a gun with kinetic energy of

20kJ. Calculate the speed of the bullet.40g = 0.04kg40g = 0.04kg

Solution:Solution:K.E. = ½ m vK.E. = ½ m v22

20 000 = 0.5 x 0.04 x v20 000 = 0.5 x 0.04 x v22

20 000 20 000 = v= v22

0.5 x 0.040.5 x 0.04Try on your own Q5.Try on your own Q5.Answer: m = Answer: m = 40kg40kg

vv2 2 = 1000, 000= 1000, 000

vv = 1000 = 1000 m/sm/s

Eg. 3

The speed-time graph of a car is shown below. The car has a mass of 1200kg.

t/h

v/kmh-1

36

00

A Ba. State the KE of the car at t=0h.K.E. = 0J

b. Find the speed of the car at A in ms-1.

Speed of car at A = 36km / h36000 60 x 60

= 10 m/s

=

c. Calculate the KE of the car at A. d. State the KE of the car at B.KE = 60 000J K.E. =60 kJ

The law of conservation of energy The law of conservation of energy states thatstates thatenergyenergycannotcannot be be createdcreated or or destroyeddestroyed but it can be but it can be transformed from one form to anothertransformed from one form to another..

Eg. 1

Fig shows an object of mass 1 kg thrown vertically upwards from the ground with an initial velocity of 4 ms-1. It reaches a maximum height of 0.8m before falling back to the ground.

a. Find its initial KE.

= = ½½ x 1 x (4)x 1 x (4)22Initial K.E.= ½ m vInitial K.E.= ½ m v22

= = 8J8J0.8m

1 kg

b. Fill in the missing information in the table below.

Height KE PE Total Energy

Remarks

0.80m 8J 8J Maximum P.E.Minimum K.E.

0.20m 2J P.E. < K.E.

0.60m

0.40m

0.00m

8J P.E. > K.E.2J

P.E. = K.E.

8J 0J Minimum P.E. Maximum K.E.

1 kg

0.8m

0J

6J

8J4J 4J

6J 8J

8J

Eg. 2 Indicate the type(s) of energy (max/min) at positions A, B, C, D and E. [Refer to textbook Page 106]

A E

B

C

DReference level

In the real world, we know that oscillating pendulum will eventually come to a halt. Why?

E.g. 3A 4.0 Kg mass of metal has fallen through a distance 0.8m onto a horizontal surface in order to test its hardness.

a) When the mass has fallen through 0.8m, how much GPE has been transformed? (g = 10N/kg)

Solution:Solution:

= 4 = 4 kgkg x 10 x 10 N/kgN/kg x 0.8 x 0.8 mm= 32 = 32 NmNm

GPE = GPE = mghmgh

= 32 = 32 JJ

b) Just before the mass hits the horizontal surface, how much KE does the mass possess?

Solution:Solution:

Gain Gain in KE = in KE = LossLoss in GPE in GPE Applying the Applying the Law of Conservation of energyLaw of Conservation of energy, ,

= 32 = 32 JJ

c) What is the velocity of the body just before hitting the horizontal surface?

Solution:Solution:KE = KE = ½ m v ½ m v 22

32 = ½ (4.0) v32 = ½ (4.0) v22

vv22 = 16= 16

v v = 4 = 4 m/sm/s

d) Assuming that 90% of the energy becomes heat energy, how much heat energy is produced when the body hits the horizontal surface?

Solution:Solution:Heat energy produced = 0.9 x 32Heat energy produced = 0.9 x 32

= 28.8 = 28.8 JJ

E.g. A frictionless metal track is curved in a shaped as shown in the diagram. A ball is released from rest at A and slides down the slope. At which pt(s)

a. is the KE greatest?b. Same as B?Is it possible for the ball to roll beyond F? Why?

A

B

CD

E

F

1m 1m

We can all complete a cross-country run. But

some take 10mins while others take 10

hours.

Power is Power is defined as the rate of doing work or the ratedefined as the rate of doing work or the rateof change of energy from one form to another.of change of energy from one form to another.

Work doneWork done Energy changeEnergy changePowerPower == ==

Time takenTime taken Time takenTime taken

p =p = WW SI unit is W or JsSI unit is W or Js--11

tt

where P is powerwhere P is power (Watts, W)(Watts, W)W is work doneW is work done (Joules, J)(Joules, J)t is time takent is time taken (sec, s)(sec, s)

E.g. 1 Fill in the blanks E.g. 1 Fill in the blanks

Force Distance moved in the direction of

force

Time taken

Work Done Power

3N 2m 4s

50N

600N

12m

5m

10s

12s

6J 1.5W

600J 60W

3000J 250W

E.g. 2A bricklayer lifts 50 bricks, each weighing 15N,

through a vertical height of 1.2m in one minute and places them at rest on a wall. Calculate

a) the work done,

Solution:Solution:

Work Done = Work Done = F x dF x d= 50 x 15 = 50 x 15 NN x 1.2x 1.2mm

= 900 = 900 NmNm= 900 = 900 J J

Solution: Power = Solution: Power = Work DoneWork DoneTime taken Time taken 900 900 JJ= = 1 1 minmin900 900 WW= =

A bricklayer lifts 50 bricks, each weighing 15N, through a vertical height of 1.2m in one minute and places them at rest on a wall. Calculate

b) the average power needed,

Power = Power = Work DoneWork DoneTime taken Time taken 900 900 JJ= =

15 15 WW= = 60 60 ss

E.g. 3A staircase in a building has 75 steps, each one 15cm. A

student of weight 500N runs up the staircase in 30s. What is the student’s average power while climbing the stairs?

Solution:Solution:

Height of 1 step = 15cm Height of 1 step = 15cm = 0.15 = 0.15 mm

Height of 75 steps, h = 75 x 0.15 = 11.25m Height of 75 steps, h = 75 x 0.15 = 11.25m

Work Done in running up 75 steps = Work Done in running up 75 steps = F x dF x d= 500= 500N x N x 11.2511.25mm= 5 625 = 5 625 J J

Average Power = Average Power = Work DoneWork DoneTime taken Time taken 5625 5625 JJ= =

187.5 187.5 WW= = 30 30 ss