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8/18/2019 2ndLE Lecture 16 - R6 Work
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Lecture 16 Objectives: Work
1. Determine the work done by constant force
acting on a system.
2. Determine the total work done on a system
composed of one or more objects.
3. Determine the work done by a varying force on a
system from a force-vs-displacement graph.
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2http:
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Work is not a job or a measure of exhaustion.
3
As a scientific concept, work is not a JOB.
You do work by exerting the force on a body while that body
moves from one place to another .
Work is related to force and displacement.
= ∙ = where is the angle between and
In 2D motion, = + and = + = +
Work has the unit of joule (J). 1 J = 1 N∙m
Recall dot product: = ∙ = ℎ
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Work can be positive, negative, or zero.
4
Case 1: Force is along the motion.
=
Work is positive.
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Case 2: A component of force is along the motion
=
Work is positive.
Work can be positive, negative, or zero.
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Case 3: Force is opposite the direction of motion.
= −
Work is negative.
Work can be positive, negative, or zero.
Negative work means work is done by the object (not on the object)
We will discuss more about this next week
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Case 4: Force is perpendicular to the direction of
motion.
= Work is zero.
Work can be positive, negative, or zero.
Summary (i f motion and forc e appl ied are in 1D):
Same direct ion o f force and d isplacement: pos i t ive
Oppo site direct ion of force and disp lacement: negativeperpendicu lar direct ion of force and displacement: zero
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What is the work done
by the lifter?
POSITIVENEGATIVE
ZERO
What is the work doneby the gravity?
POSITIVE
NEGATIVE
ZERO
Work is done in lifting the
barbell.
If the barbell could be lifted
twice as high, the weight lifterwould have to do twice as
much work.
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Work is done in lifting the
barbell.
If the barbell could be lifted
twice as high, the weight lifterwould have to do twice as
much work.
What is the work done
by the lifter?
POSITIVENEGATIVE
ZERO
What is the work doneby the gravity?
POSITIVE
NEGATIVE
ZERO
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Given:
F = 210N
s = 18m
angle = 30oRequired: Work
Sample Problem: Work done by a constant force
Steve exerts a steady force of magnitude 210N on a
stalled car as he pushes it a distance of 18m. The carhas a flat tire, so to make the car track straight Steve
must push at an angle of 30o to the direction of motion.
How much work does Steve do?
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Given: =160 −40 = 14 + 11
Required: Work
Sample Problem: Work done by a constant force
Steve pushes a second stalled car with a steady
force =160 −40. The displacement of thecar is =14 +11 . How much work doesSteve do in this case?
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Work can also be done by multiple forces.
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Work done by each force just add up.
= ∆ + 2∆ + ⋯ = + 2 + ⋯ ∆ = ∆
If all of forces cancel out, total work done on thesystem is zero; however, work done by each forcemay not be zero.
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Given:
F or FT = 5000N
d = 20mw = 14700N
f = 3500N
Angle = 36.9o
Required: work done by all forces acting and WTotal
Sample Problem: Work done by several forces
A farmer hitches her tractor to a sled loaded with firewood and
pulls it a distance of 20m along level ground. The total weight of
sled and the load is 14700N. The tractor exerts a constant 5000-N force at an angle of 36.9o above the horizontal. There is a
3500-N friction force opposing the sled’s motion.
Find the work done by each force acting on a sled and the total
work done by all the force.
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Given:
F or FT = 5000Nd = 20m
w = 14700N
f = 3500N
Angle = 36.9o
Required: work done by all forces acting and WTotal
Wf ,WT, Ww Wn
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Wn = 0 and Ww = 0since motion is perpendicular
to these forces
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Summary
16
= ∙ = where is the angle between and
REMINDER:
RECIT 6 (DUE: FRIDAY)
PROBLEM SETS (WRITE YOUR
SECTION ABOVE YOUR NAME)
EXAM RESULTS (LATER - WED)
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= + − = −2
= −
= + − = + 2
= +
At the top, the equation of motion is:
At the bottom, the equation of motion is:
(1)
(2)
(3)
(4)
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Seatwork- solve problems in your
notebooks
- write the answers only inyour bluebook
- indicate the date
March 01, 20161. Blah?
2. Blah blah!
3. Blah blah blah!
4. Blah blah blah blah!
18
1 L li h 10 0 k bi t h i t ll l l f
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1. Leslie pushes a 10.0 kg cabinet horizontally on a level surface
with an applied force of 25.0 N. If the cabinet slides on the surface
by 5.00 m, what is the work done by gravity on the cabinet?
(a) - 491 J (b) - 125 J (c) 0 J (d) 125 J
2. A loaded grocery cart is rolling across a parking lot in a strong
wind. You apply a constant force = 30.0N − 40.0N to the cartas it undergoes a displacement = −9.0m − 3.0m . How muchwork did you exert on the grocery cart?
(a) 150J (b) 390J (c) 150J (d) -390J
A factory worker pushes (horizontally) a 30.0kg crate with a distance of
4.5m along a level floor at constant velocity. The coefficient of kinetic
friction between the crate and floor is 0.25. (Draw FBD)3) What magnitude of force did the worker apply?4) How much work is done on the crate by the worker’s force?
5) How much work is done on the crate by the friction?
6) How much work is done on the crate by the normal force?
7) How much work is done on the crate by the gravitational force?8) Calculate the total work done on the crate?