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Work

The document explains the concept of work in physics, defined as the product of force and displacement when the force is applied in the same direction as the movement. It includes formulas for calculating work, examples of problems involving work done by forces, and the conditions under which work is performed. Additionally, it discusses the units of measurement for force and work.

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rosieteheinrich
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11 views25 pages

Work

The document explains the concept of work in physics, defined as the product of force and displacement when the force is applied in the same direction as the movement. It includes formulas for calculating work, examples of problems involving work done by forces, and the conditions under which work is performed. Additionally, it discusses the units of measurement for force and work.

Uploaded by

rosieteheinrich
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PPTX, PDF, TXT or read online on Scribd
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 the change in position of an object.

 Vector
WORK
C
Work is done when an object is
moved by a force.

Work done as the product of the


force and the distance the object
travels (W=Fd) when the force is
constant and applied in the same
direction the object moves.
Work done No Work done
dot product

Two vectors depicting their


direction and described as the

𝑾 = 𝑭⃗ ∙ 𝒅⃗= |𝑭||𝒅|
product of force and
displacement
When a force acts upon an object to cause
a displacement of the object, it is said that
work was done upon the object. Other
than that, work can only be done on an
object if a component or all of the force
exerted on it is parallel or antiparallel to
the object’s displacement; they must not
be perpendicular.
For the vector 𝒅⃗ , we can represent it
in a straight line from one point to

force applied 𝑭⃗ to an object. Let’s say


another point while having a constant

a displacement from point A to point B


as the object moves:

W = 𝑭⃗· 𝑨𝑩⃗ =|𝑭⃗|·|𝑨𝑩⃗|cos θ = Fd cosθ


UNITS

F (Force) = newton (N) or 𝑘𝑔.m/s2


Where

W (Work done) = J (Joule) or 𝑘𝑔.m2/s2


d (displacement) = meter (m)
Calculating Work
the projection of 𝑭⃗ onto the positive x-axis.
Sample Problem 1: The horizontal component of the force is

A boy is pulling a wagon


with a force having a magnitude of
10N on the handle at an angle
of 55°. If the boy pulls the wagon 60
meters, find the work done by the
force.
Given:
F = 10N
d = 60m
θ = 55°
Required: W=?
Formula: W=Fd cosθ
Solution: W=Fd cosθ
= (10N) x (60m) cos(55°)
= (10N) x (60m) x 0.5736
Answer: = 344.15 N· m or
= 344.15 J
Sample Problem 3.
A 25-N force is applied to push a block
across a frictional surface at a constant
speed for a displacement of 10.0 m to the
right. Calculate the work done by the applied
force.
Given: Fapp = 25N
d = 10m
θ = 0°
Required: Work = ?
Formula: W = Fd cosθ
Solution: = (25N) x (10m) cos(0°)
= (25N) x (10m) x 1
= 250 N· m
Answer: = 250 J
Given: Ff = -25N
d = 10m
θ = 0°
Required: Work = ?
Formula: W = Fd cosθ
Solution: = (-25N) x (10m) cos(0°)
= (-25N) x (10m) x 1
= -250 N· m
Answer: = 250 J
A PORTER PULLS A 10 KG LUGGAGE ALONG A
LEVEL ROAD FOR 5 M BY EXERTING A FORCE OF
20 N AT AN ANGLE OF 300 WITH THE
HORIZONTAL SHOULDER THROUGH A VERTICAL
DISTANCE OF 1.5 M AND CARRIES IT FOR
ANOTHER 5 M. hOW MUCH WORK DOES HE DO
IN (A) PULLING, (B) LIFTING AND (C) CARRYING
THE LUGGAGE ON HIS SHOULDER?

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