Work and Energy

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Work and Energy
Work
Physics definition of Work:
Work : is the product of the magnitudes of the
component of force along the direction of
displacement and the displacement
W = work
F = force
D = displacement
W = Fd
F=ma
W=mad
Work
displacement
force
WORK!
displacement
Work is done only when
components of a force are
parallel to a displacement.
NO WORK!
force
Is Work Happening?
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A tug of war that is evenly matched
A student carries a bucket of water along a
horizontal path while walking at a constant
velocity.
A Crane lifting a car.
A person holding a heavy chair at arm’s
length for several minutes.
A train engine pulling a loaded boxcar initially
at rest.
Work Units
Work = F x d
Work = m x a x d
Work = (newtons) (m)
(Newton x m) = joules (J)
Work Problem 1
A tugboat pulls a ship with a constant horizontal
net force of 5.00 x 103 N. How much work is
done if the ship is pulled a distance of 3.00
km?
W=Fxd
= 5.00 x 10 3 N ( 3000 m)
= 1.5 x 107 Nm or J
Work Problem 2
If 2.0 J of work is done raising a 180 g
apple, how far is it lifted?
W =Fd
F = mg = 0.18kg(9.8 m/s2) = 1.76 N
W = Fd
2 J = (1.76N)d
d = 1.1 m
Work Problem 3
A weight lifter lifts a set of weights a vertical
distance of 2.00 m. If a constant net force of
350 N is exerted on the weights, what is the net
work done on the weights?
W = Fd
W = 350 N x 2.00 m = 700 Nm or 700J
Sample Problem 4
What work is done by a forklift raising 583 kgs
of frozen turkeys 1.2 m?
W = Fd
F = 583 kg (9.8 m/s2) = 5713.4 N
W = 5713.4 N ( 1.2m) = 6856 Nm or 6856 J
Problems with Forces at
Angles
F
θ
X-direction
Fx = Fcos θ
Fy = Fsin θ
Because the displacement of the box is only in the
x direction only the x-component of the force does work
on the box.
W = Fdcosθ
Sample Problem
A sailor pulls a boat a distance of 30.0 m along
a dock using a rope that makes an angle of
25o with the horizontal. How much work is
done if he exerts a force of 255 N on the
rope?
255 N
250
W = Fdcosθ
= 255N(30m)cos25
= 6.93 x 103J
Sliding up an Incline
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What we calculated was...
For sliding an object up an incline..
W = Fd
W = (mg sinθ) d
Sample Problem
An airline passenger carries a 215 N suitcase
up stairs, a displacement of 4.20 m vertically
and 4.60 m horizontally. How much work
does the passenger do?
4.2 m
4.6 m
Sample Problem
First have to calculate
hypotenuse and θ
6.23 m
4.2 m
42.40
Tan θ = 4.2 m/ 4.6 m
Θ = 42.40
Hypotenuse2 = A2 + B2
Hypotenuse2 = (4.2)2 + (4.6)2
Hypotenuse = 6.23 m
4.6 m
Sample Problem
Suitcase weighs 215 N
4.2 m
42.40
4.6 m
F║ = mg sinθ = 215 sin 42.4 = 145 N
W = Fd = 145 N ( 6.23 m) = 899 Nm = 903 J
This should equal the force of the suitcase moving it
vertically 4.2 m W = 215 N ( 4.2 m) = 903 J
Force
Force
Graphs of Force vs.
Displacement
Displacement
Displacement
Work = Fd
Work can be found graphically by finding the
area under the curve
Homework
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Do Work/Energy/Power worksheet #1-4
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