Download Work Non-Isolated Systems

Work
Work
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Specific Outcomes:
i. I can describe, qualitatively, the change in mechanical energy in a system
that is not isolated.
ii. I can recall work as a measure of the mechanical energy transferred and
power as the rate of doing work.
iii. I can solve, quantitatively, kinematics and dynamics problems, using the
work-energy theorem.
iv. I can use free-body diagrams to organize and communicate solutions to
work-energy theorem problems.
Work
Non-Isolated Systems
Work-Energy Theorem
Work at Angles
Work
Non-Isolated Systems
If energy is transferred to an object, work is
done on that object (work = energy)
Remember: for an isolated system, the
overall mechanical energy is constant
If a force is applied over a distance, work is:
Energy is not lost or gained by the system, as
long as only conservative forces are involved
where:
W = Fd
W = work or energy (J)
F = force applied (N)
d = distance travelled (m)
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
If non-conservative forces are involved,
energy must be added to the system or
removed from it
The Law of Conservation of Energy applies!
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
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Non-Isolated Systems
Work is done:
by the surroundings on the system, or
Non-Isolated Systems
Since work changes the overall mechanical
energy (Em) of the system:
W = ∆Em
by the system on the surroundings
If the mechanical energy of the system
increases, work has been done on the system
If the mechanical energy of the system
decreases, work has been done on the
surroundings
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work (W) is the change in energy (∆Em)
Work can be positive (+), if the surroundings
do work on the system
Work can be negative (-), if the system does
work on its surroundings
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Non-Isolated Systems
Here, initial mechanical energy (Em1) and
final mechanical energy (Em2) are not the
same, since work is done:
Non-Isolated Systems
Case #1: a model rocket burns its fuel to launch
straight up
Em2
Em2 = Em1 + W
Em2 = final mechanical energy of
the system (J)
Em1 = initial mechanical energy of
the system (J)
W = system’s change in energy (J)
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
energy
where:
Ep
Em1
Ek
time
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
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Non-Isolated Systems
Here, the system is the rocket
Non-Isolated Systems
Case #2: a ball drops against air friction
Work is done on the system by the fuel being
burned
Em2
energy
∆Em is positive here, since the final
mechanical energy (Em2) is greater than the
initial mechanical energy (Em1)
Em1
Ep
Ek
time
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Non-Isolated Systems
Here, the system is the ball
Work is done on the surroundings (air) by the
system (falling ball)
∆Em is negative here, since the initial
mechanical energy (Em1) is greater than the
final mechanical energy (Em2)
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Non-Isolated Systems
If there are multiple sources of work, we may
use the following (if needed):
Em2 = Em1 + (WA + WB + WC + … + Wf)
The sign on each work term depends on
whether it adds or subtracts from the overall
mechanical energy of the system
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
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Work-Energy Theorem
Work-energy theorem - the work done on a
system is equal to the sum of the changes in
potential and kinetic energies of the system:
W = ∆Ek + ∆Ep
In this equation:
∆Ek = Ekf – Eki
∆Ep = Epf - Epi
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work-Energy Theorem
ex. A 1.5 kg block is pushed up an inclined plane. At the
bottom, the block had a height of 0.50 m and a
speed of 2.00 m/s. At the top, the block has a
height of 3.50 m and a speed of 4.50 m/s. Calculate
the work done to move the block.
∆Ep = Epf - Epi
= mghf – mghi
= (1.5 kg)(9.81 m/s2)(3.50 m)
– (1.5 kg)(9.81 m/s2)(0.50 m)
= +44.145 J
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work-Energy Theorem
ex. A 1.5 kg block is pushed up an inclined plane. At the
bottom, the block had a height of 0.50 m and a
speed of 2.00 m/s. At the top, the block has a
height of 3.50 m and a speed of 4.50 m/s. Calculate
the work done to move the block.
∆Ek = Ekf - Eki
= ½mvf2 – ½mvi2
= ½(1.5 kg)(4.50 m/s)2 – ½(1.5 kg)(2.00 m/s)2
= +12.1875 J
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work-Energy Theorem
ex. A 1.5 kg block is pushed up an inclined plane. At the
bottom, the block had a height of 0.50 m and a
speed of 2.00 m/s. At the top, the block has a
height of 3.50 m and a speed of 4.50 m/s. Calculate
the work done to move the block.
W = ∆Ek + ∆Ep
= (12.1875 J) + (44.145 J)
= 56 J
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
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Work at Angles
Remember that a force may be applied at an
angle to a horizontal surface:
Work at Angles
We can find the work done on an object
when force is applied at an angle:
where:
Fapp
θ
W = work done on the object (J)
We used trigonometry to find the parallel
component:
Fapp
F = force applied at an angle (N)
θ = angle of the applied force (°)
d = distance traveled (m)
=
θ
Fapp
W = Fd cosθ
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work at Angles
ex. A desk is pushed forward at an angle of 30° to the
floor at a constant speed by a force of 50 N. Find
the work done by moving the desk 25.0 cm.
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
Work at Angles
ex. A force of 10 N [176°] is applied to a toy car,
making it travel 20.0 cm [102°]. Determine the
work done on the object.
d = 25.0 cm ÷ 100 = 0.250 m
d = 20.0 cm ÷ 100 = 0.200 m
W = Fd cosθ = (50 N)(0.250 m)(cos30)
W = Fd cosθ = (10 N)(0.200 m)(cos74)
= 11 J
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
= 0.55 J
Dulku – Physics 20 – Unit 3 (Circular Motion, Work and Energy) – Topic K
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