Section 6.1 Mechanical Work

Section 6.1 Mechanical work • Mechanical work done • Energy

© Manhattan Press (H.K.) Ltd.

1

6.1 Mechanical work (SB p. 121)

Mechanical work How can we describe one’s ability in doing things?

Work HOW? ~!

Go to © Manhattan Press (H.K.) Ltd.

Quiz

2

6.1 Mechanical work (SB p. 121)

Mechanical work Work has been Work done on bodies

• a force is applied to make something move / try to stop it from moving

• there is energy transfer © Manhattan Press (H.K.) Ltd.

3

6.1 Mechanical work (SB p. 122)

Mechanical work done mechanical work done – measure of energy transfer

W=F× s • W – work done • F – applied force • s – the displacement along the direction of F

Thinking 1

© Manhattan Press (H.K.) Ltd.

4

6.1 Mechanical work (SB p. 122)

Mechanical work done e.g. F F

More work is done

Work done (W) — scalar quantity — unit: joule (J)

© Manhattan Press (H.K.) Ltd.

5

6.1 Mechanical work (SB p. 122)

Force at angle θ to displacement

W = Fs cos θ F F cos θ

© Manhattan Press (H.K.) Ltd.

6

6.1 Mechanical work (SB p. 123)

No work done 1. Body is stationary W=0 2. Moving with its own inertia 3. Applied force perpendicular to displacement 2. v ≠ 0, F = 0

uniform velocity

1. v = 0

3. F ⊥ s to right

mg mg

Thinking 2 © Manhattan Press (H.K.) Ltd.

Thinking 3 7

6.1 Mechanical work (SB p. 123)

Example 1: Find the work done on a box of mass 8 kg in each of the following cases. (a) The box slides on a stationary frictionless conveyer belt at uniform velocity over a distance of 2 m. Solut ion

Since no force is applied in the direction of displacement of the box, W = 0. The box moves because of its own inertia. © Manhattan Press (H.K.) Ltd.

8

6.1 Mechanical work (SB p. 124)

Example 1: (Cont) (b) A man lifts up the box through a vertical distance of 1.5 m. To lift the box, the man must apply a force that is equal to the weight of the box. Solut F=m× g ion = 8 × 10 = 80 N Since the applied force and the displacement are in the same direction (upward), work done is: W=F× s = 80 × 1.5 = 120 J © Manhattan Press (H.K.) Ltd.

9

6.1 Mechanical work (SB p. 124)

Example 1: (Cont) (c) He carries the box and walks through a horizontal Solut distance of 3 m. ion Since the direction of the applied force is perpendicular to that of the displacement, the angle is 90°. Force in the direction of displacement of box is: F = mg cos 90° =0N ∴ W=0J

© Manhattan Press (H.K.) Ltd.

10

6.1 Mechanical work (SB p. 124)

Class Practice 1: A force of 60 N is applied and the suitcase moves through a distance of 10 m in each case. Calculate the work done. (a) The force is applied in the same direction as the motion of the suitcase. Ans wer F×s W = _________________ 60 × 10 = _________________ 600 J = _________________ © Manhattan Press (H.K.) Ltd.

11

6.1 Mechanical work (SB p. 124)

Class Practice 1 (Cont): (b) The force is applied at an angle of 30° to the Ans direction of motion of the suitcase. wer Fs cosθ W = _________________ 60 × 10 × cos30° = _________________ 520 J = _________________

© Manhattan Press (H.K.) Ltd.

12

6.1 Mechanical work (SB p. 125)

Energy Energy

• Capacity to do work • Measured in Joule (J) sleeping

swimming © Manhattan Press (H.K.) Ltd.

walking

running 13

6.1 Mechanical work (SB p. 125)

Energy Energy exists in many forms, e.g. : Making sound Heating

Moving © Manhattan Press (H.K.) Ltd.

14

6.1 Mechanical work (SB p. 125)

Energy PE

h

Energy can exist in different forms potential energy kinetic energy + (PE or Ep) (KE or Ek)

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

15

To section 6.2

© Manhattan Press (H.K.) Ltd.

16

6.1 Mechanical work (SB p. 120)

Quiz

1. At which point does the durian posses the highest potential energy? A. A © Manhattan Press (H.K.) Ltd.

Ans wer 17

6.1 Mechanical work (SB p. 120)

Quiz (Cont)

2. At which point does the durian posses the highest kinetic energy? C. C © Manhattan Press (H.K.) Ltd.

Ans wer 18

6.1 Mechanical work (SB p. 120)

Quiz (Cont)

3. A watermelon is hold still by Sally, does she do any work from the Ans scientific point of view? wer B. No. Since the watermelon does Return to not move. Text

© Manhattan Press (H.K.) Ltd.

19

6.1 Mechanical work (SB p. 122)

Thinking 1:

Refer to Fig. 6.3. Find the work done by the force on the object if the F–s graph is as shown. Ans wer

The area under the F-s graph is the work done. W = Fs = 5 × 6 = 30 J © Manhattan Press (H.K.) Ltd.

Return to

Text 20

6.1 Mechanical work (SB p. 123)

Thinking 2:

In the following cases, are the people doing any work from the scientific point of view? Ans wer Fig. (a) Sally is sitting still to read a book

No work has been done. No force is applied and there is no movement. © Manhattan Press (H.K.) Ltd.

21

6.1 Mechanical work (SB p. 123)

Thinking 2 (Cont): Fig. (b) Louis is holding a dumbbell Ans wer

No work has been done. He applies a force but there is no movement.

© Manhattan Press (H.K.) Ltd.

22

6.1 Mechanical work (SB p. 123)

Thinking 2 (Cont): Don’t just stand there! Do some work!

What do you mean I’m not doing work?

Her comment is justified. Louis is not doing work. © Manhattan Press (H.K.) Ltd.

Fig. (c) If the door does not move, do you think the comment by Sally is justified? Ans wer Return to

Text

23

6.1 Mechanical work (SB p. 124)

Thinking 3:

What a player hits a billiard with a snooker, he makes the contact distance of the snooker with the billiard as long as possible. Explain Ans briefly. wer

The player tries to make the force acting on the billiard for a longer distance. So, greater work is done on the billiard for it to gain more energy. Return to © Manhattan Press (H.K.) Ltd.

Text

24