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Work 1

The document explains the concept of work in physics, defining it as the product of force and displacement in the direction of the force, and emphasizes that work is a scalar quantity measured in Joules. It outlines different scenarios for calculating work based on the direction of force and displacement, including cases where they are parallel, opposite, perpendicular, or at an angle. Several examples illustrate the calculation of work in various contexts, such as gravitational force and lifting objects.
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0% found this document useful (0 votes)
17 views32 pages

Work 1

The document explains the concept of work in physics, defining it as the product of force and displacement in the direction of the force, and emphasizes that work is a scalar quantity measured in Joules. It outlines different scenarios for calculating work based on the direction of force and displacement, including cases where they are parallel, opposite, perpendicular, or at an angle. Several examples illustrate the calculation of work in various contexts, such as gravitational force and lifting objects.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPTX, PDF, TXT or read online on Scribd
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Work

APS ACADEMY
What is Work?
 When a force acts on an object and makes it
move, we say that the force has done work on
the object.
 In this figure, gravity is making the
ball move downwards.

Thus it is doing work on the ball


 The work done by a force is given by:
Work is a Scalar Quantity
 Work = Force × Displacement in the direction of
force

Force and displacement are both vector quantities.

However, work is a scalar quantity.

i.e., work does not have any direction, it is


specified only by the magnitude.
 The unit of Work is Joule.
Case of Different directions
of Displacement and Force
 There are four cases possible:

1. Displacement is in the direction of the applied force


2. Displacement is in the opposite direction of the applied
force
3. Displacement is perpendicular to the applied force
4. Displacement is at an angle ‘θ’ to the force

The standard formula is given by:

Work done = Force × Displacement × cos(θ)


Different Directions of
Force and Displacement
1. Displacement and force are in the same direction

Work done = Force × Displacement

Because cos(θ) = 1 for θ = 0˚

2. Displacement and force are in the opposite direction

Work done = – Force × Displacement

Because cosθ = –1 for θ = 180˚


Different Directions of
Force and Displacement
3. Displacement and Force are perpendicular

Work done = 0

Because cos(90˚) = 0

In this figure, as the aeroplane moves horizontally,


the work done by gravity is 0 because it is acting
perpendicularly
Different Directions of
Force and Displacement
 Displacement and Force are at an angle ‘θ’

Work done = Force × Displacement × cos(θ)

For example, if the angle between


them is 45˚

Work = (F×D)/√2
Example
 Find the work done by Earth on Moon in
completing one revolution.

Given: distance between Earth and Moon = 384000


km

Mass of Earth = 6×1024 kg

Mass of Moon = 7.3×1022 kg


Example contd.
 The direction of moon is
perpendicular to the
gravitational force exerted by
Earth.

Therefore, work done by Earth


on Moon = 0.
Example
 A 1kg ball thrown upwards reaches a maximum
height of 5m. Find the work done by gravity
during this vertical displacement.
Example contd.
The force of gravity = weight of the ball = 1×9.8
= 9.8N

Displacement is given as 5m

However, they are in opposite directions,


because displacement is upwards and gravity
acts downwards.
 Work done = 9.8×5 = –49J
Example
 An aeroplane of mass 1000kg moves from
Delhi to Gurgaon (10km).
 Find the work done by gravity.
Example contd.
 Work done by gravity = 0 Joule
Example
 A roller is pushed with a force of 100N along its
handle, which is at an angle of 30˚ with the
horizontal. Find the work done in moving it
through 10m.
 Given: cos(30˚) = √3/2
Example contd.
Work done = F*s*cos(θ)

Force = 100N; angle = 30˚; s = 10m

 Work done = 100*10*cos(30˚) = 1000*√3/2 =


500√3
Example
 Calculate the work done by a student in lifting a
0.5kg book from the ground and keeping it on a
shelf 1.5m high.
Example contd.
 Work done = mgh

= 0.5*9.8*1.5

= 7.35N
Example
 Calculate the work done in lifting 200kg of
water through a vertical height of 6 metres.
Example contd.
Work done = mgh

= 200×10×6

= 12000J
Example
 A car weighing 1000kg and travelling at 30m/s
stops at a distance of 50m decelerating
uniformly.
 What is the force exerted on it by the brakes?

 What is the work done by the brakes?


Example contd.
 Force = ma

Given: u = 30m/s; v = 0m/s; s = 50m

Using v2 = u2 + 2as

We get a = –9m/s

Since F = ma
 F = 1000×(-9) = –9000N

 Work done = F×s = –9000×50 = –450,000 Joules


Example
 A porter lifts a luggage of 15kg from the
ground and puts it on his head 1.5m above
ground.
 Calculate the work done by him on the
luggage.
Example contd.
 Force exerted by porter = 15×9.8 = 147N

Vertical displacement = 1.5m


 Work done = 147×1.5

= 220.5 Joules
Example
 A man weighing 70kg carries a weight of
10kg on the top of a tower 100m high.
 Calculate the work done by the man.
Example contd.
 Force exerted by the man = (70+10)×9.8 =
784N

Vertical displacement = 100m


 Work done = 784×100

= 78400Joules
Example
 A porter carries a load of 50kg on his head
and walks on a level road upto 100m.
 What is the work done by him?
Example contd.
 Work done = 0 Joules
Example
 A boy of mass 55kg runs up a flight of 40
stairs, each measuring 0.15m.
 Calculate the work done by the boy.
Example contd.
 Force exerted by the boy = 55×9.8 = 539N

Displacement = 40×0.15 = 6m
 Work done by man = 539×6

= 3234 Joules
Example
 An engine pulls a train 1km over a level track.

 Calculate the work done by the train given that


the frictional resistance is 5×105N.
Example contd.
 Frictional resistance = F = 5×105 N

Work done by frictional force = – 5×105 ×103

= –5×108 Joule
Force

θ Displacement

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