Section 6.4 Law Of Conservation Of Energy

Section 6.4 Law of conservation of energy

• Law of conservation of energy • Inter-conversion of PE and KE • Inter-conversion of PE and KE with energy loss © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 131)

Law of conservation of energy Energy exists in many forms, e.g. : Sound energy Heat energy

Kinetic energy

Can we transform the forms of energy? Yes! © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 131)

Law of conservation of energy KE of water PE of water KE of Turbines



Electrical energy light, heat and other forms of energy

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6.4 Law of conservation of energy (SB p. 132)

Law of conservation of energy Law of conservation of energy — energy cannot be created / destroyed — can be transformed from one form to another — total amount remains unchanged thermal energy, E1

electrical energy, E E = E1 + E2 + E3 + E4 © Manhattan Press (H.K.) Ltd.

light energy, E2 sound energy, E3

kinetic energy, E4

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6.4 Law of conservation of energy (SB p. 132)

Class Practice 4: 1. State the energy transfer in each of the following situations. The first one is done as an example. (a) A boy lifts a dumbell upwards.

The chemical energy released in the muscle of the boy changes to the potential energy of the dumbbell. © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 132)

Class Practice 4 (Cont): (b) A player hold a bat to hit a ball. Ans wer

The chemical energy of the player changes to the KE of the ball, with some dissipated as heat and sound.

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6.4 Law of conservation of energy (SB p. 132)

Class Practice 4 (Cont): (c) A student throws a basketball into the net. Ans wer

The chemical energy of the student changes to the PE and KE of the basketball. © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 132)

Class Practice 4 (Cont): (d) A windmill generates electricity from wind. Ans wer

The wind energy changes to the KE of the blades and turbines, and then to the electric energy.

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6.4 Law of conservation of energy (SB p. 133)

Class Practice 4 (Cont): (e) Use a magnifying glass to light up a match. Ans wer

The solar energy changes to heat and light.

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6.4 Law of conservation of energy (SB p. 133)

Class Practice 4 (Cont): (f) Launch a spaceship from the ground. Ans wer

The chemical energy of the fuel changes to heat, sound, PE and KE of the spaceship. © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 133)

Class Practice 4 (Cont): 2. In the case of striking a match with a matchbox, what causes the energy changes? Ans wer The work done against the frictional force during striking causes the chemical energy of the match to be changed to heat and light.

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6.4 Law of conservation of energy (SB p. 133)

Inter-conversion of PE and KE In free-falling motion

Workshop 1 Conservation of energy

PE decreases

h

CAL

How How are they they related? related? KE increases

Attraction from earth © Manhattan Press (H.K.) Ltd.

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Activity 1 Discussion 2

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6.4 Law of conservation of energy (SB p. 135)

In free-falling motion Loss in PE = Weight × Distance = mgs By

v2 = u2 + 2as v2 = 0 + 2gs 1 1 ( m )v2 = ( m)2gs 2 2 1 2 mv = mgs 2

Gain in kinetic energy = Loss in potential energy © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 135)

In free-falling motion Gain in kinetic energy = Loss in potential energy • Energy loss due to air resistance (is neglected in this case) PE + KE = constant PE and KE are interchangeable

Thinking 5 © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 136)

Example 3: A ball of mass 2 kg is released from rest at position A. Take PE at the ground as zero. Find the potential energy and kinetic energy of the ball Solut (a) at position A, and ion (b) at position B. (b) PEB = mghB (a) PEA = mghA = 2 × 10 × 6 = 120 J KEA = 0 J © Manhattan Press (H.K.) Ltd.

= 2 × 10 × (6 − 1) = 100 J By the law of conservation of energy, Gain in KE at B = Loss in PE from A to B KEB = 120 − 100 = 20 J

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6.4 Law of conservation of energy (SB p. 136)

Class Practice 5: A ball is thrown vertically upwards at a speed of 16 m s−1. Find the maximum height the ball can reach. Ans Gain in PE = Loss in KE wer 1 2 mgh = 2 mv v2 h = 2g 162 = 2 × 10 = 12.8 m © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 137)

In pendulum motion Examples of pendulum:

swing © Manhattan Press (H.K.) Ltd.

bob of grandfather clock 17

6.4 Law of conservation of energy (SB p. 137)

In pendulum motion Expt 6A Energy conversion in a simple pendulum

A

B weight

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C

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6.4 Law of conservation of energy (SB p. 138)

In pendulum motion CAL Workshop 2 Conservation of potential and kinetic energy

speed increases

From A to B: loss in PE = gain in KE © Manhattan Press (H.K.) Ltd.

speed decreases

Thinking 6

From B to C: loss in KE = gain in PE 19

6.4 Law of conservation of energy (SB p. 139)

Example 4: A pendulum bob is released from rest at point A as shown in Fig. (a). Assume the air resistance is negligible. Find the speed of the bob at the lowest point B. Solut ion

Fig. (a) © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 139)

Example 4: (Cont) Let the speed of the bob at B be v (Fig. (b)). Loss in PE = Gain in KE 1 mgh1 = mv22 1 m × 10 × (1 − cos30°) = 2 mv2

∴

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v = 1.64 m s−1

Fig. (b)

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6.4 Law of conservation of energy (SB p. 139)

Class Practice 6: Refer to Example 4. If the bob is moving at a speed of 1 m s−1 at point A, find the maximum height that the bob can reach on the other side. Let the bob reach the highest point C on the other side and let the height be h2 from point B. By the law of conservation of energy, Ans wer

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6.4 Law of conservation of energy (SB p. 139)

Class Practice 6 (Cont): PE at C = KE at A + PE at A 1 mgh2 = 2 mu2 + mgh1 1 m × 10 × h2 = 2 m × 12 + m × 10 × (1 − cos30°) ∴ h2 = 0.18 m

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6.4 Law of conservation of energy (SB p. 140)

Inter-conversion of PE and KE with energy loss In reality, swinging pendulum’s height is decreasing gradually.

energy loss by friction © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 140)

Inter-conversion of PE and KE with energy loss

s f

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Loss in PE = Gain in KE + Work done against friction

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6.4 Law of conservation of energy (SB p. 140)

Inter-conversion of PE and KE with energy loss Work done against friction = fs (if f is constant) m

s

s

f m

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friction ( f )

Thinking 7

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6.4 Law of conservation of energy (SB p. 140)

Example 5: Jessie of mass 30 kg sits on the top of a slide of height 3 m. The slide board is 8 m long. She slides down and reaches the bottom of the slide at a speed of 2.5 m s−1.

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6.4 Law of conservation of energy (SB p. 141)

Example 5: (Cont) Find Solut (a) the work done against friction, and ion (b) the average friction acting on Jessie. (a) Work done against friction = PE at the top − KE at the bottom 1 2 W = mgh − mv 2 1 = 30 × 10 × 3 − × 30 × 2.52 2

= 900 − 93.8

Solut (b) ion

806 = f × 8 ∴

101 N

= 806 J

W =fs f

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= 28

6.4 Law of conservation of energy (SB p. 141)

Class Practice 7: A ball of 5 kg is released from point A of height h from the ground. It moves along the track to reach the ground at point B with a speed of 10 m s−1. Given that the total work done against friction between the ball and the track is 50 J. Find h. Ans wer

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6.4 Law of conservation of energy (SB p. 141)

Class Practice 7 (Cont):

Loss in PE = Gain in KE + Work done against friction 1 mgh = mv2 + 50 2 1 5 × 10 × h = × 5 × 102 + 50 2 ∴ h =6m

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To section 6.5

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6.4 Law of conservation of energy (SB p. 133)

Activity 1: Conservation of energy Let’s start: 1. Set up the apparatus on a table as shown. A weight is hung by a thread to act as a pendulum bob. 2. Place a block beside the weight and mark the position of the block.

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6.4 Law of conservation of energy (SB p. 133)

Activity 1 (Cont): Conservation of energy

3. Pull the weight to one side and measure its height (h) from the table. Release the weight and let it hit the block. Measure the distance travelled by the block (d). Repeat the experiment by releasing the weight from different heights. Record the readings below. h / cm d / cm

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6.4 Law of conservation of energy (SB p. 133)

Activity 1 (Cont): Conservation of energy

State the relation between h and d. Explain briefly.

Ans wer

The higher the position of the weight to be released, the longer the distance travelled by the block. This indicates that more energy is transferred from the weight to the block. © Manhattan Press (H.K.) Ltd.

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Text 34

6.4 Law of conservation of energy (SB p. 134)

Discussion 2:

Describe the changes in PE and KE for the diver jumping from the springboard to the water as shown. Ans wer

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6.4 Law of conservation of energy (SB p. 134)

Discussion 2 (Cont): When the diver stands on the end of the spring, elastic PE is stored in the spring. When she jumps, elastic PE is released and changes to KE of the diver. Then she moves upwards and KE changes gradually to PE. At the highest point, all KE is changed to PE. Then she falls because of the pull of gravity and PE changes to translational and rotational KE again. Finally when she jumps into the water, KE is changed into heat and sound, and workdone against the resistance of water (the most). Return to © Manhattan Press (H.K.) Ltd.

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6.4 Law of conservation of energy (SB p. 136)

Thinking 5:

An object falls freely at a certain height above the ground. Sketch graphs to show the variations of 1. KE, PE and ME with distance fallen, and 2. KE, PE and ME with time 1 1. and 2. KE + PE = ME 1. 2.

Ans wer

PE = mgh = mg ( 2 gt2) 1 mg2 t2 = 2 PE ∝ t2 Return to

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Text 37

6.4 Law of conservation of energy (SB p. 138)

Thinking 6:

In Fig. 6.16, when the weight swings, work is done on the weight by the tension in the string. Is this correct? Explain briefly. Ans swings,wer

When the weight tension on the string is always perpendicular to the direction of motion. Therefore, no work is done on the weight. © Manhattan Press (H.K.) Ltd.

T

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6.4 Law of conservation of energy (SB p. 140)

Thinking 7:

Release a basketball from a height of about 1 m. Describe its motion and state whether there is energy loss during each rebound of the ball. The ball released from rest moves downwards. It reaches the highest speed before colliding with the ground and it changes it directions after the collision. It rebounds with a decreasing speed and rises up to a certain point. After each rebound, the basketball rises to a smaller height than before. So its mechanical energy decreases. The energy is lost as heat, sound and the work done against friction. © Manhattan Press (H.K.) Ltd.

Ans wer

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