Energy and Work

• both are scalars --- easier to manipulate

• introduce new physical principle:conservation of energy

Kinetic Energy state of motion

• If speed << c, then

relative motion fixed frame of reference

joule N∙m

Work transfer of energy

• work done by a force on a system• energy to system +W energy from system −W• no material transfer• • • if constant, • • • units same as energy

φ

��

��

Work transfer of energy

when then and K↑when then and K↓when then and K is constant

Work transfer of energy

when then and K↑when then and K↓when then and K is constant

F

d

Work transfer of energy

when then and K↑when then and K↓when then and K is constant

d

F

Work transfer of energy

when then and K↑when then and K↓when then and K is constant

d

F

Work and Kinetic Energy

• work done on system

•

Work Done by Gravity

• is always down, varies.

Fg

d

d

d

d

Fg

0°

180°φφ

Work Done by Gravity

• is always down, varies.

• with applied force Fa

• • when then •

Fg

d

d

d

d

Fg

0°

180°φφ

dφ h

consider vertical drop h

𝑊 𝑔= h𝑚𝑔 cos0 °= h𝑚𝑔

in triangle h=𝑑cos𝜑

therefore, substituting in above

𝑊 𝑔=𝑚𝑔𝑑 cos𝜑

An 8.00 kg cat tumbles 2.00 m into a pool. How much work was done by gravity on the cat?

Mighty Mouse tumbles in after the cat. How fast was he moving on impact with the water?

A champion curler must apply what constant force at a 15° angle to a 15.0 kg stone over a 1.50 m distance on frictionless ice to give it a final velocity of 1.10 m/s?

A pitcher releases a 145 g baseball at 40.0 m/s. How much work must be done by the pitcher on the ball?

What constant force must he apply to the ball over a 2.10 m distance?

The catcher’s mitt moves back 10.0 cm in stopping the ball. How much work was done by the catcher on the ball? What constant force?

l = 1.5 m

h = 0.91 m

A worker pushes a 45kg block of ice up a frictionless ramp at a constant speed. Determine the applied force, the work done by the applied force, the work done by gravity, the work done by the normal force, and the net work.

a

tt0 t1

area = at

a = a

𝑎=𝑑𝑣𝑑𝑡 𝑑𝑣=𝑎𝑑𝑡

∫𝑡 0

𝑡 1

𝑑𝑣=∫𝑡 0

𝑡 1

𝑎𝑑𝑡=𝑎∫𝑡 0

𝑡 1

𝑑𝑡

𝑣−𝑣0=𝑎 ( 𝑡1− 𝑡0 )=𝑎𝑡

𝑣=𝑣0+𝑎𝑡

Integration anti-derivative

Integration anti-derivative

v

tti tf

v = v0 + at

½at2

viv0t

𝑣=𝑑𝑦𝑑𝑡 𝑑𝑦=𝑣 𝑑𝑡

∫𝑡 0

𝑡 1

𝑑𝑦=∫𝑡0

𝑡1

𝑣 𝑑𝑡=∫𝑡 0

𝑡 1

(𝑣0+𝑎𝑡 )𝑑𝑡

𝑦− 𝑦0=∫𝑡 0

𝑡1

𝑣0𝑑𝑡+∫𝑡0

𝑡1

𝑎𝑡 𝑑𝑡

𝑦− 𝑦0=𝑣0∫𝑡 0

𝑡 1

𝑑𝑡+𝑎∫𝑡 0

𝑡 1

𝑡 𝑑𝑡

𝑦− 𝑦0=𝑣0 𝑡+½ 𝑎𝑡2

𝑦=𝑦0+𝑣0 𝑡+½𝑎𝑡2

area = v0t + ½at2

Work Done by a Varying Force

Particle acted on by a varying force. Clearly, ·d is not constant!F

Work Done by a Varying Force

For a force that varies, the work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up.

Work Done by a Varying Force

In the limit that the pieces become infinitesimally narrow, the work is the area under the curve:

Or:

Work Done by a Varying Force

Work done by a spring force:

The force exerted by a spring opposes displacement and is given by Hooke’s Law:

The constant k is the spring or elastic constant

Work done by a spring force:

Fs

x

Fs = −kx

kx𝑊 𝑠=∫

𝑥0

𝑥1

[𝐹 𝑠(𝑥 )�� ]∙ [𝑑𝑥�� ]=∫𝑥0

𝑥1

𝐹 𝑠 (𝑥 ) 𝑑𝑥

𝑊 𝑠=∫𝑥0

𝑥1

−𝑘𝑥 𝑑𝑥

Work done by applied force: 𝑊 𝑎=½𝑘𝑥2

¿½ 𝑘𝑥𝑖2−½ 𝑘𝑥𝑓

2

¿½ 𝑘𝑥𝑓2 −½𝑘𝑥 𝑖

2

Work done by spring: 𝑊 𝑠=−½𝑘 𝑥2|𝑖𝑓

In preparation for lab Week 8, read Ch. 14, sect. 1-6, of our text: Oscillatory Motion.

Don’t be left hanging from a spring.

A yearly tradition at MIT is the water balloon fight between dorm buildings. Latex tubing is attached to the window frames to act as sling shots. Assume the tubing follows Hooke’s Law with a spring constant 114 N/m. If the tubing is initially stretched 2.60 m, find the work done on the balloon by the latex tubing (Ws). Find the launch velocity of the water balloon, mass 0.500 kg).

Projectile Launcher

xi xf

vivf

𝐹 𝑠=−𝑘𝑥 𝑊 𝑠=−½𝑘𝑥2¿½ 𝑘𝑥𝑖

2−½ 𝑘𝑥𝑓2

Cannot use: this F is only at one x.

Mass of ball: 60.0 gxi = 12.0 cmxf = 0 cmvi = 0 m/svf = 4.90 m/s

Find the spring constant k.

Projectile Launcher

x i

x f

v i

v f

mass of ball: 60.0 gxi = 12.0 cmxf = 0 cmvi = 0 m/sθ = 30.0°Ws = 0.720 Jk = 100. N/m

Calculate the launch velocity.θ

Projectile Launcher

xi

xf

vi

vf

Same conditions but now with an angle of 90°

What is the launch velocity?

Potential Energy

• due to configuration, arrangement, position• Gravitational: position in gravitational field• Elastic: stretch/compress around a position• U = 0 the choice is yours

• Potential Energy the result of forces, the interaction of bodies; not property of the object: object does not “have” potential energy.

• Potential Energy always involves a conservative force

Conservative Forces

• if work done by force depends on initial/final positions and NOT the path

• if initial = final position then Wnet = 0, no matter what path …. Wforward = Wreverse

• gravitational & spring/elastic forces are conservative

• Friction (including drag) is nonconservative:» total Wfric < 0 never ≥ 0

» path dependent

Potential Energy and Work

• Gravitational

•

•

•

Potential Energy and Work

• Elastic (follows Hooke’s Law)

•

•

Conservation of Mechanical Energy

• No frictional forces permitted

• • /

Work done on a system

𝑊=∆𝐸=∆𝐸 h𝑚𝑒𝑐 +∆𝐸𝑁𝐶

for an isolated system (no E or mass transfer)

∆𝐸 h𝑚𝑒𝑐 +∆𝐸𝑁𝐶=0

When unloaded the spring of the launcher reaches the muzzle. When a 100 g ball is placed in the launcher the spring is now compressed by 1.10 cm.What is the spring constant?

When unloaded the spring of the launcher reaches the muzzle. When a 100 g ball is placed in the launcher the spring is now compressed by 1.10 cm.What is the spring constant?When the launcher is fully armed the spring has been compressed an additional 11.0 cm. Just before firing what is the elastic potential energy of the system with respect to the unloaded launcher?

When unloaded the spring of the launcher reaches the muzzle. When a 100 g ball is placed in the launcher the spring is now compressed by 1.10 cm.What is the spring constant?When the launcher is fully armed the spring has been compressed an additional 11.0 cm. Just before firing what is the elastic potential energy of the system with respect to the unloaded launcher?

After firing the ball is propelled upwards. At its maximum height what is its potential energy with respect to the launch position?

When unloaded the spring of the launcher reaches the muzzle. When a 100 g ball is placed in the launcher the spring is now compressed by 1.10 cm.What is the spring constant?

𝑘=98.0𝑁 /𝑚When the launcher is fully armed the spring has been compressed an additional 11.0 cm. Just before firing what is the elastic potential energy of the system with respect to the unloaded launcher? 𝑈 𝑠=0.717 𝐽

After firing the ball is propelled upwards. At its maximum height what is its potential energy with respect to the launch position?∆𝐾 +∆𝑈=0=𝐾 𝑓 −𝐾 𝑖+𝑈 𝑠𝑓 −𝑈 𝑠𝑖+𝑈𝑔𝑓 −𝑈𝑔𝑖

0 − 0 + 0 − 0.717J + Ugf − 0

𝑈𝑔𝑓=0.717 𝐽

When unloaded the spring of the launcher reaches the muzzle. When a 100 g ball is placed in the launcher the spring is now compressed by 1.10 cm.What is the spring constant?

𝑘=98.0𝑁 /𝑚When the launcher is fully armed the spring has been compressed an additional 11.0 cm. Just before firing what is the elastic potential energy of the system with respect to the unloaded launcher? 𝑈 𝑠=0.717 𝐽

After firing the ball is propelled upwards. At its maximum height what is its potential energy with respect to the launch position? 𝑈𝑔𝑓=0.717 𝐽

What is the ball’s maximum height above the launch position?

A pendulum cat swings from an initial displacement angle of 10.0° at the end of a 20.0 m line. At the bottom of the swing a knife edge severs the cord just above the cat. What is the cat’s velocity at the break?

5.00 m

8.00 kg

θ

A

A pendulum cat swings from an initial displacement angle of 10.0° at the end of a 20.0 m line. At the bottom of the swing a knife edge severs the cord just above the cat. What is the cat’s velocity at the break?How far to the right of point A does the cat land?

5.00 m

8.00 kg

θr

h

l

A

A pendulum cat swings from an initial displacement angle of 10.0° at the end of a 20.0 m line. At the bottom of the swing a knife edge severs the cord just above the cat. What is the cat’s velocity at the break?How far to the right of point A does the cat land? What is the tension in the line at the bottom?

5.00 m

8.00 kg

θr

h

l

A

h

A

B

r

Design for a frictionless roller coaster in Newton’s amusement park. Find the ratio of h to r such that at B there is no normal force.

In the Neonewton Saloon (friction allowed) a full beer stein of mass 2.00 kg slid horizontally across the bar with an initial speed of 3.00 m/s. The stein travelled a distance of 1.53 m before stopping (without losing a drop). Find the coefficient of kinetic friction between the stein and the bartop.