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Work and Energy

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Work and Energy

Work is the measure of energy transfer .

- Define kinetic energy and potential energy, along with the appropriate units.
- Describe the relationship between work and kinetic energy, and apply the WORK-ENERGY THEOREM.
- Define and apply the concept of POWER.

Energy

Energyis anything that can be converted into work; i.e., anything that can exert a force through a distance. The SI unit for energy is the joule (J)

Energy is the capability for doing work.

Potential Energy:Ability to do work by virtue of position or condition.

The SI unit for energy is the joule (J)

Potential energy is measured as a difference from an arbitrarily chosen reference point

Gravitational Potential Energy

What is the P.E. of a 50-kg person at a height of 480 m?

U = mgh = (50 kg)(9.8 m/s2)(480 m)

U = 235 kJ

- Kinetic Energy: Ability to do work by virtue of motion. (Mass with velocity)
- Examples â€“
- A speeding car
- A bullet

5 g

200 m/s

What is the kinetic energy of a 5-g bullet traveling at 200 m/s?

What is the kinetic energy of a 1000-kg car traveling at 14.1 m/s?

vf

d

vi

F

F

m

m

A force acting on an object changes its velocity, and does work on that object.

The Work-Energy Theorem: The work done by a force is equal to the change in kinetic energy that it produces.

Work is equal to the change inÂ½mv2

If we define kinetic energy as Â½mv2 then we can state a very important physical principle:

6 cm

0

80 m/s

d

F = ?

Example 1: A 20 g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s.

Work = Â½mvf2 - Â½mvi2

F d= - Â½mvi2

F (0.06 m) cos 1800 = -Â½ (0.02 kg)(80 m/s)2

F = 1067 N

F (0.06 m)(-1) = -64 J

Work to stop bullet = change in K.E. for bullet

d=25m

f

DK =Â½mvf2 - Â½mvi 2

vi= 2mkgd

vi = 2(0.7)(9.8 m/s2)(25 m)

Example 2: A bus slams on brakes to avoid an accident. The tread marks of the tires are 25m long. If mk = 0.7, what was the speed before applying brakes?

Work = DK

Work =F(cos q) d

F= mk.FN= mkmg

Work = - mkmg d

-Â½mvi 2 = -mkmgd

vi = 18.52 m/s

Kinetic Energy:Ability to do work by virtue of motion. (Mass with velocity)

Potential Energy:Ability to do work by virtue of position or condition.

The Work-Energy Theorem: The work done by a force is equal to the change in kinetic energy that it produces.

Work = Â½mvf2 - Â½mvi2

Law of conservation of energy â€“ the total energy of a system remains unchanged unless an outside force does work on the system, or the system does work on something outside the system.

The energy of a system can interconvert between potential and kinetic energy, but the sum of the two will always remain constant.

Sample problem1 : A 5.0 kg ball is dropped from a

height of 10.0 meters.

a. What is its velocity when it is 8.0 meters high?

b. What is its velocity immediately prior to hitting

the ground?

Solution:

a. Before it is dropped the ball only has

gravitational potential energy since there is no motion.

Therefore:

E=U+K=U+0=mgh=(10)(5)(9.81)= 490.5J=U

- Solution (continued):
- (continued). At 8.0 meters the ballâ€™s potential energy
- is U=(8)(5)(9.81)=392.4J
- Since its total energy is conserved,
- E=U+K and K=E-U
- K=490.5-392.4=98.1J

Solution:

b. Right before the ball hits, h=0, U=0,

and E=K+0=490.5J

Pendulum â€“ the energy of a pendulum is also

a result of work done against gravity. The pendulum

is at equilibrium when the bob hangs straight down.

As the bob is moved from equilibrium it rises and

acquires potential energy.

Pendulum â€“ the potential energy of a pendulum

displaced at an angle qfrom equilibrium is

U=mgh = mgl(1-cos q)

Where l is the length of the pendulum

The energy of the system E is

E=mgl(1-cos qmax)

Where qmaxis the angle at the maximum displacement

from equilibrium.

As the pendulum swings it loses potential energy

and gains kinetic energy. At the bottom of the swing

U=0 and K=E

- Example:
- A pendulum is 2.0m long, and has a bob that has a
- mass of 2.0 kg. The bob is pulled from the
- equilibrium position, so that the angle with the
- vertical axis Î¸=10o, and released.
- What is the potential energy of the system before
- the bob is released?
- b. What is the speed of the bob as it goes
- through the equilibrium position?
- c. What is the speed of the bob when the angle
- with the vertical is 5.0o?

- Example:
- What is the potential energy of the system before
- the bob is released?
- U=mgl(1-cosqmax)=E=(2)(9.81)(2)(1-cos10o)=0.59J

Example:

b. What is the speed of the bob as it goes

through the equilibrium position?

At equilibrium, U=0, and E=K

Example:

c. What is the speed of the bob when the angle

with the vertical is 5.0o?

K=E-U

U=mgl(1-cos q)=(2)(9.81)(2)(1-cos5.0o)=0.149J

K=E-U=0.59J-0.149J=.441J

After the mass is released, the spring stretches or compresses the velocity of the mass changes and the energy of the system interconverts between potential and kinetic energy.

The maximum displacement from the equilibrium position is called the amplitude A.

Example:

A system consists of a spring attached to a 1.0kg mass

that slides on a frictionless horizontal surface.

The spring has a force constant of 200N/m.

a. If the spring is stretched to a distance of 0.20m from

the equilibrium position, what is the potential energy of

the system?

b. What is the speed of the mass when it is going through

the equilibrium position?

c. What is the speed of the mass when it is at a

distance of 0.1 m from the equilibrium position?

Example:

a. If the spring is stretched to a distance of 0.20m from

the equilibrium position, what is the potential energy of

the system?

Potential energy = the work done on the system

Example:

b. What is the speed of the mass when it is going through

the equilibrium position?

At the equilibrium position, U=0 and K=E

Example:

c. What is the speed of the mass when it is at a

distance of 0.1 m from the equilibrium position?

At 0.1m from equilibrium E=U+K

During an elastic collision â€“ when two moving objects collide, the kinetic energy of the system is the same before and after the collision.

During an inelastic collision â€“ when two moving objects collide, the kinetic energy of the system is not the same before and after the collision.