Unit Three: EnergyChapter 4 and 6

Work

p. 181 1-7 extra p. 183 1-7 p. 226 1-4, 9-11Booklet p.23 1-4 p.24 13, 14, 18, 19, 24, 26, 30

Kinetic Energy and the Work-Energy Theoremp. 186 1,2,4-8 extra p. 188 1-8 p. 226 5,6,12,13Booklet p.23 5-12 15-17, 20-23, 25, 27-29

Work

The work done by a constant force is defined as the product of the component of the force in the direction of the displacement and the magnitude of the displacement.

dFW

cos dFW

W = Work units of J or Nm or

the Joule is named after James Prescott Joule

is the angle between the force and the displacement

2

2

s

mkg

Note that Fcos is the component of the force in the direction of the displacement. If the angle is greater than ninety degrees then the work will be negative (cos.

F

d

F

cosF

Work is a scalar quantity. Energy is defined as the ability to do work and therefore is a scalar quantity as well. Work can be positive or negative but these signs are not direction. We will see that they indicate a gain of kinetic energy or a loss of kinetic energy respectively.

Negative work is done on an object when it is slowed by a force. Positive work is done when an object is sped up by a force.

The area under a F-d graph is equal to the work done by an applied force. Assume the force and displacement are colinear.

The total work done on an object is the sum of all the work done by individual forces.

dFW

dFW R

Rousseau pushes with a force of 500 N on an immovable wall. How much work is done on the wall?

David swings a rock around his head with a centripetal force of 250 N. The rock goes around his head 3 times in 0.56 s (the radius of the circle is 0.8 m). What is the work done on the rock?

A 4 kg block is raised 5 m. How much work is done on the block?

aF

gF

assume constant v

JW

ms

mkgW

dmgW

dFW g

2.196

)1)(5)(81.9)(4(

0cos

cos

2

workF a

JW

mkg

NkgW

dmgW

dFW g

2.196

)1)(5)(81.9)(4(

180cos

cos

workF g

This means the work done on the ball is 0 J.

note

JW

W

dFW net

0

)5)(0(

A sled (15 kg) is pulled with a 50 N [20o ath] force for 7.5 m. The coefficient of friction is 0.21. Calculate the work done by each force and total work done on the sled.

NF

aF

gF

fF

20o

Work done by FN and Fg are zero since they are perpendicular to the displacement.

Work done by Fa

JW

mNW

dFW a

4.352

)20)(cos5.7)(50(

cos

][0.130

)81.9)(15(20sin50

0

upNF

kg

NkgF

FFF

FFF

N

N

gayN

gayN

To calculate work done by friction we must calculate FN.

JW

W

dFW

dFW

N

f

8.204

)1)(5.7)(0.130)(21.0(

180cos

cos

Work done by Ff

Therefore the total work done on the sled is 147.6 J

WORK ENERGY THEOREM

For an object that is accelerated by a constant net force and moves in the same direction . . .

cos)(

cos

cos

12 dt

vvmW

dmaW

dFW net

22

)(2

)2

)((

)1()(

21

22

21

22

1212

12

mvmvW

vvm

W

vvvvmW

t

dvvmW

define kinetic energy as (Ek)

2

2mv

k

kk

EW

EEW

12

The work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

22cos

21

22 mvmv

dF

A ball is dropped from rest at a height of 8.25 m. What will be its speed when it hits the ground? (could solve this kinematically but let’s do it using the work-energy theorem.

22cos

22cos

22

22

12

12

mvmvdmg

mvmvdF

EW

g

k

divide by m

gF

s

mv

vm

s

m

72.12

02

)1)(25.8)(81.9(

2

2

2

2

Therefore the speed of the ball is 12.72 m/s when it hits the ground.