Motion, forces and energy - work
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21 Terms
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work
is the product of force and the distance moved in the direction of the force.
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what are the SI units for work
Joules (J)
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what are the SI units for force
Newtons (N)
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what are the SI units for distance
Metres (m)
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formulae for work
Work = force x distance
W = F X d
W = F X d
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NOTE
- the distance MUST be in the direction of the force
- the distance must be converted into metres
- the distance must be converted into metres
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what is a kilojoule (1 KJ)
1000 joules or 10^3 joules
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what is a megajoule (1 MJ)
1,000,000 joules or 10^6 joules
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what is a gigajoule (1 GJ)
1,000,000,000 joules or 10^9 joules
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work done in raising an object
in lifting the object, the minimum force required is equal to the object
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formulae for work done in raising an object
mass x gravitational pull x height
m x g x h
m x g x h
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NOTE
mg --> force
h --> distance
h --> distance
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Work done on an inclined plane
- input force = force x length (work done by effort)
- output/useful work done = m x g x h
- wasted work done = work done against friction
= friction force x length
= input work - output work
- output/useful work done = m x g x h
- wasted work done = work done against friction
= friction force x length
= input work - output work
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QUESTIONS
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1. A 1200kg car is pulled 12m up an incline by a rope exerting a force of 6000N. The car experiences a 1000n friction force.
(a) determines useful work
(a) determines useful work
w = m x g x h
= 1200 x 10 x 5000
= 60,000,000
= 60 MJ
= 1200 x 10 x 5000
= 60,000,000
= 60 MJ
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1.(b) calculate input work
input w = F x l
= 6000 x 12
= 72,000J
= 6000 x 12
= 72,000J
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1.(c) calculated wasted work
wasted w = Fr x l
= 1000 x 12
= 12,000J
= 1000 x 12
= 12,000J
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2. How much work is done by a force of 1.0 N moving through 1m
w = F x d
= 1 x 1
= 1J
= 1 x 1
= 1J
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3. How much work is done by a force of 5.0 N moving through 2m
w = F x d
= 5 x 2
= 10J
= 5 x 2
= 10J
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4. Which does more work, a force of 500N moving 10m or a force of 100N moving 40m
w = F x d
= 500 x 10
= 5000 J
w = F x d
= 100 x 40
= 4000 J
ans: a force of 500N moving 10m
= 500 x 10
= 5000 J
w = F x d
= 100 x 40
= 4000 J
ans: a force of 500N moving 10m
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5. Three boxes each weigh 100N. A man lifts all the boxes together from the ground in 2.0 s to do this. Calculate the useful work the man does to lift the boxes.
w = m x g x h
m = weight ÷ 10
m = 100 ÷ 10
m = 10
w = m x g x h
= 10 x 10 x 1.5
= 150
1 box = 150J
3 boxes = 150 x 3
= 450J
m = weight ÷ 10
m = 100 ÷ 10
m = 10
w = m x g x h
= 10 x 10 x 1.5
= 150
1 box = 150J
3 boxes = 150 x 3
= 450J
Note 
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