2ndle lecture 16 - r6 work

8/18/2019 2ndLE Lecture 16 - R6 Work

1/19

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.


2/19

2http:

//betterexplained.com/articles/vector-calculus-understanding-the-dot-product/

ab

ax

ay by

bx


3/19

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


4/19

Work can be positive, negative, or zero.

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Case 1: Force is along the motion.

=

Work is positive.


5/19

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Case 2: A component of force is along the motion

=

Work is positive.



6/19

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Case 3: Force is opposite the direction of motion.

= −

Work is negative.


Negative work means work is done by the object (not on the object)

We will discuss more about this next week


7/19

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Case 4: Force is perpendicular to the direction of

motion.

= Work is 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


8/19

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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.


9/19

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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


10/19

10

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?


11/19

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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?


12/19

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.


13/19

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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.


14/19

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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

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= ∙ = where is the angle between and

REMINDER:

RECIT 6 (DUE: FRIDAY)

PROBLEM SETS (WRITE YOUR

SECTION ABOVE YOUR NAME)

EXAM RESULTS (LATER - WED)


17/19

= + − = −2

= −

= + − = + 2

= +

At the top, the equation of motion is:

At the bottom, the equation of motion is:

(1)

(2)

(3)

(4)


18/19

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


19/19

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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?

2ndle lecture 16 - r6 work

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