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ASPhysicsCIERevision Notes5. Work, Energy & Power5.1 Energy ConservationWork & Energy

Work & Energy (CIE AS Physics)

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Leander

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Leander

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Work done

  • In physics, work is done when an object is displaced by an external force applied in the direction of its displacement

W space equals space F s

  • Where:
    • W = work done in newton metres (N m) or joules (J)
      • 1 N m = 1 J
    • F = force in newtons (N)
    • s = displacement of object in metres (m)

 

  • In the diagram below, the man’s pushing force on the block is doing work as it is transferring energy to the block (increasing its kinetic energy)

Man pushing a block

Work done diagram, downloadable AS & A Level Physics revision notes

Work is done when a force is used to move an object over a distance

 

  • When work is done, energy is transferred 
  • Work done is equal to the amount of energy transferred
  • Usually, if a force acts in the direction that an object is moving, then the object will gain energy
  • If the force acts in the opposite direction to the movement, then the object will lose energy

 

Worked example

The diagram shows a barrel of weight 2.5 × 103 N on a frictionless slope inclined at 40° to the horizontal.

WE - Work done on barrel question image, downloadable AS & A Level Physics revision notesA force is applied to the barrel to move it up the slope at constant speed.

The force applied is parallel to the slope.

How much work is done in moving the barrel a distance of 6.0 m up the slope?

A.     7.2 × 103 J               B.     2.5 × 104 J              C.     1.1 × 104 J               D.     9.6 × 103 J

Answer: 

Step 1: List the known quantities

  • Weight, FW = 2.5 × 103 N at 40°
    • It is helpful to use FW for the force of weight since W is used for work done
  • displacement, s = 6.0 m

Step 2: State the work done equation

W space equals space F s

Step 3: Calculate the force in the direction of travel

  • The force, FA, applied to push the barrel needs to overcome the component of the barrel's weight that is parallel to the slope
  • Since the force is parallel to the slope, the component of weight to consider is that which is parallel to the slope

5-1-1-we-work-done-on-barrel-answer-cie-new

F subscript A space equals space F subscript W space cross times space sin open parentheses 40 close parentheses

F subscript A space equals space 2.5 cross times 10 cubed space cross times space sin open parentheses 40 close parentheses

F subscript A space equals space 1607 space straight N

Step 4: Substitute the known values into the work done equation to calculate

W space equals space 1607 space cross times space 6.0

W space equals space 9.6 cross times 10 cubed space straight J

  • Therefore, the correct answer is D

Examiner Tip

A common exam mistake is choosing the incorrect force which is not parallel to the direction of movement of an object. You may have to resolve the force vector to find the component that is parallel. The force may not always be in the same direction as the movement, as shown in the worked example.

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    Leander

    Author: Leander

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.