Work March 18, 2013 - p. 1/9

March 18, Week 9

Today: Finish Chapter 6 and begin Chapter 7, Energy

Homework Assignment #6 - Due Friday, March 22

Mastering Physics: 9 problems from chapters 5 and 6

Written Questions: 6.73

Exams and Midterm grades are in your mailbox. Exam #2grade is on white sheet.

If interested in Physics 110, you can still start attendingtomorrow

Help sessions with Jonathan:

M: 1000-1100, RH 111 T: 1000-1100, RH 114Th: 0900-1000, RH 114

Work March 18, 2013 - p. 2/9

Work Review

Work - How much effort goes into causing motion.

Unit: N ·m = kg ·m2/s2 = J .

Work March 18, 2013 - p. 2/9

Work Review

Work - How much effort goes into causing motion.

Unit: N ·m = kg ·m2/s2 = J .

−→

F

−→sφ

−→s = displacement= distance and direction φ = angle between

−→

F and −→s

W = Fs cosφ =−→

F ·−→s

For Constant Force and Straight-line Motion:

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

x

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

x

F

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

s

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

s

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

s

Variable Force

F

x

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

s

Variable Force

F

x

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

s

Variable Force

F

xx1 x2

s

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

sW = area

Variable Force

F

xx1 x2

s

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

sW = area

Variable Force

F

xx1 x2

s

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

sW = area

Variable Force

F

xx1 x2

s

W = area

Work March 18, 2013 - p. 3/9

Variable Forces

To find the work done by a changing force, we have to find thearea under a curve.

Constant Force, W = Fs

F

xx1 x2

F

F

sW = area

Variable Force

F

xx1 x2

s

W = area

For any type of force, it can be shown that the work-energy

theorem holds! Wtotal = ∆K = 1

2mv22 −

1

2mv21

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Pav =∆W

∆t

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Pav =∆W

∆tunit: J/s = Watt

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Pav =∆W

∆tunit: J/s = Watt

P = lim∆t→0

∆W

∆t=

dW

dt=

−→

F ·−→v

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Pav =∆W

∆tunit: J/s = Watt

P = lim∆t→0

∆W

∆t=

dW

dt=

−→

F ·−→v

In the U. S., unit of work is lb · ft. The unit of power should bethe lb · ft/s, but we use the horsepower (hp).

Work March 18, 2013 - p. 4/9

Power

Power - The rate at which work is done.

Pav =∆W

∆tunit: J/s = Watt

P = lim∆t→0

∆W

∆t=

dW

dt=

−→

F ·−→v

In the U. S., unit of work is lb · ft. The unit of power should bethe lb · ft/s, but we use the horsepower (hp).

1hp = 550 lb · ft/s = 746Watt

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

(c) 2m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

(c) 2m/s

(d) 3m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

(c) 2m/s

(d) 3m/s

(e) 4m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

(c) 2m/s

(d) 3m/s

(e) 4m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

(a) 0m/s

(b) 1m/s

(c) 2m/s

(d) 3m/s

(e) 4m/s

Work March 18, 2013 - p. 5/9

Power Exercise

The power supplied by the person pulling the rope can’texceed the power of the rising block! If the person is pulling therope down at 4m/s, with what speed is the block rising?

PHand = T (4m/s) (Downwards pulland velocity ⇒ φ = 0◦)

PBlock = (4T )vblock (Upwards forceand velocity on block ⇒ φ = 0◦ heretoo)

PHand = PBlock ⇒ T (4m/s) = (4T )vblock

(b) 1m/s

Work March 18, 2013 - p. 6/9

Potential Energy

Some forces do work that can be saved or stored.

Work March 18, 2013 - p. 6/9

Potential Energy

Some forces do work that can be saved or stored.

Potential Energy, U - Saved or stored energy, i.e., energy thatcan be converted into kinetic energy at a later time.

Most textbooks define potential energy as energy that dependson position. That is true for the examples we do in physics, butnot true in every case.

Work March 18, 2013 - p. 6/9

Potential Energy

Some forces do work that can be saved or stored.

Potential Energy, U - Saved or stored energy, i.e., energy thatcan be converted into kinetic energy at a later time.

Most textbooks define potential energy as energy that dependson position. That is true for the examples we do in physics, butnot true in every case.

Conservative Forces - Forces that create potential energy.

Conservative forces are rare. Only gravity and the spring forceare conservative. (You’ll learn two more next term - the electricand magnetic force.) For a force to be conservative, the work itdoes must be independent of path.

Work March 18, 2013 - p. 7/9

Conservation of Energy

For a conservative force,

W = −∆U

Work March 18, 2013 - p. 7/9

Conservation of Energy

For a conservative force,

W = −∆U

Conservation of Energy - If only conservative forces do workon an object, its total energy cannot change.

Total Energy, E = the sum of kinetic and potential energy.

E = K + U

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

∆K = −∆U

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

∆K = −∆U

K2 −K1 = − (U2 − U1)

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

∆K = −∆U

K2 −K1 = − (U2 − U1) ⇒ K1 + U1 = K2 + U2

Work March 18, 2013 - p. 8/9

Conservation of Energy II

Proof: If a conservative force is the only force doing work on anobject then:

Wtotal = W

The work-energy Theorem ⇒ Wtotal = ∆K.

−∆U

∆K = −∆U

K2 −K1 = − (U2 − U1) ⇒ K1 + U1 = K2 + U2

⇒ E1 = E2

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

(b) 7.5 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

(b) 7.5 J

(c) 15 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

(b) 7.5 J

(c) 15 J

(d) 30 J

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

(b) 7.5 J

(c) 15 J

(d) 30 J

(e) Cannot be determined

Work March 18, 2013 - p. 9/9

Energy Conservation Exercise

A block having 15 J of gravitational potential energy is droppedfrom rest. When the block hits the ground, it has 15 J of kineticenergy. If gravity is the only force acting on the block, howmuch potential energy does the block have when it hits theground?

Ug = 15 J

K = 15 J

(a) 0 J

(b) 7.5 J

(c) 15 J

(d) 30 J

(e) Cannot be determined

K1 + U1 = K2 + U2 ⇒

0 + 15 J = 15 J + U2