Quiz question #1

quiz question #2

quiz question #3

Handout question #3

Mechanical Energy

Kinetic Energy: the energy of an object in motion

depends on mass and velocity

Ek = 1

2

mv2

http://www.youtube.com/watch?v=aCdHEWWpWj8&feature=related

What is the kinetic energy of a 1000 kg car traveling at 9m/s?

Ek= 1/2 (1000kg)(9m/s)2

= 40500J

Work Energy theorem

Work done = Kinetic Energy (end) -Kinetic Energy (start)

Wtot = ΔEk = Ekf - EkiWtot = FR Δd cos θ

Ek = 1 mv2

2

1/2 mvi

2 1/2 mvf

2

-Wtot =

or

A car is travelling at 15 m/s, it weighs 600kg

What is the kinetic energy?

Ek = 1 mv2

2

Ek = 0.5(600kg)(15m/s)2

= 67500J

p. 344 Example B

A dog is pulling a 25.0 kg sled over a horizontal surface of 4.00 m with a force of 50.0N using a horizontal rope. The sled was initially at rest and we assume that friction is negligible.

a) What is the total work done?b) What is the sled's final velocity?

Data: m= 25.0 kgF = 50.0N

θ = 00

Vi = 0

Δd = 4.0 mWtot = ?Vf = ?

W tot = WFT + WFg + WFN

Wtot = 50.0N 4.0m = 200J

Eki = 0 (vi = 0)W tot = Ekf

200J = 1/2m(vf)

2

400J/25.0kg= vf

2

v

f = 4.00m/s

a)

b)

How much work is needed to accelerate a 1000kg car from 20 to 30 m/s?

M = 1000kgvi = 20m/svf = 30 m/s

W=?

W = 1/2mvf

2 - 1/2mvi

2

W= 1/2 (1000kg)(30m/s)

2 - 1/2(1000kg)(20m/s)

2

W = 2.5 x 10

5J

Gravitational potential energy

Epg = mgh

Epg - Gravitational potential energy (J)m mass of object

g gravitational acceleration (m/s2)

h is height above the reference level (usually the ground) expressed in metres (m)

h= 2.0m

m = .25kg

Epg = mgh = .25kg(9.8m/s2)(2.0m)= 4.9J

Conservation of mechanical energy

Mechanical Energy = Kinetic Energy + Potential EnergyEm = Ek + Ep

Eki + Epi = Ekf + Epf

A 1000kg roller coaster car moves from point A to point B and then to point C

What is the gravitational PE at B and C relative to point A?

rollercoaster

0m

40m30m

10m

A B

C

D

The law of conservation of total energy states that the total quantity fo energy is constant in an isolated system.

If there is friction part of the mechanical energy is transformed into thermal energy.

∴ Eki + Epi = Ekf + Epf + ∆Eth

∆Eth = the final thermal energy or heat

ExampleEstimate the kinetic energy and the velocity required for a 70 kg pole vaulter to pass over a bar 5.0m high. assume the vaulter's height at the level of the bar itself.Δd = 5.0 m - 0.9m = 4.1m

1/2 mvi

2 + 0 = 0 + mgy2

= (70kg)

(9.8m/s2)(4.1)

= 2.8 x 103 J

The velocity is

1/2mvi2 = 2.8 x 10

3J

vi2 =(2/70kg)2.8 x 10

3J

vi = 8.9 m/s