Mechanical Energy

Total Mechanical Energy

Total Mechanical Energy (ET): Energy can be transferred or transformed,

never lost Law of Conservation of Energy

kgT EEE *If friction negligible

If friction is not negligible then….

FrictionkgT EEEE

A 56kg diver runs & dives from the edge of a cliff into the water which is located 4m below. If she is moving at 8m/s the instant she leaves the cliff, determine the speed at which she enters the water.

Before Jump At Water Level

Eg Ek Eg Ek

m = 56 kg

g = 9.8 N/kg

h = 4 m

m = 56 kg

v = 8 m/s

m = 56 kg

g = 9.8 N/kg

h = 0 m

m = 56 kg

v = ?

kgT EEE kgT EEE =

TT EE

Before Jump At Water Level

kgT EEE kgT EEE

)4)(8.9)(56(gE

JEg 2.2195

2)8)(56(2

1kE

JEk 1792

JET 2.3987

)0)(8.9)(56(gE

JEg 0

kE02.3987

JEk 2.3987

2))(56(2

12.3987 v

smv /93.11

A child throws a 0.2kg rock at a tree. When the rock leaves the child’s hand, it is moving at 20m/s & is located 1.5m above the ground. How high above the ground does the rock strike the tree if it is moving at 10m/s at that instant?

Throw Tree

Eg Ek Eg Ek

m = 0.2 kg

g = 9.8 N/kg

h = 1.5 m

m = 0.2 kg

v = 20 m/s

m = 0.2 kg

g = 9.8 N/kg

h = ?

m = 0.2 kg

v = 10 m/s

kgT EEE kgT EEE =

TT EE

Throw Tree

kgT EEE kgT EEE

)5.1)(8.9)(2.0(gE

JEg 94.22)20)(2.0(

2

1kE

JEk 40

JET 94.42

2)10)(2.0(2

1kE

JEk 10

1094.42 gE

JEg 94.32

)(?)8.9)(2.0(94.32 mh 81.16

A 2 kg ball rolls along a frictionless surface. It passes point C at a speed of 20m/s. What was the speed of the ball at point A?

Point A Point CEg Ek Eg Ek

m = 2 kg

g = 9.8 N/kg

h = 20 m

m = 2 kg

v = ? m/s

m = 2 kg

g = 9.8 N/kg

h = 25 m

m = 2 kg

v = 20m/s

kgT EEE kgT EEE =

A C

20 m25 m

TT EE

Point A Point C

kgT EEE kgT EEE

)20)(8.9)(2(gE

JEg 392

)25)(8.9)(2(gE

JEg 490

JEk 400

2)20)(2(2

1kE

JET 890

kE392890

JEk 498

2))(2(2

1498 v

smv /32.22

A force of 16N is applied to a 400g mass, starting at rest, over a distance of 5m. How long does it take the mass to move 5m?

Before After 5m

Eg Ek Eg Ek

Eg = 0 m = 0.4 kg

v = 0 m/s

Eg = 0 m = 0.4 kg

v = ?

Recall, work = energy

So work applied to move mass 5m

Law of Conservation

of Energy

dFW )5)(16(appliedW

JWapplied 80

Work required to set block in

motion

initialfinal kkappliedapplied EEEW

22 )0)(4.0(2

1))(4.0(

2

180 v

smv /20

atvv 12

t)40(1020

advv 221

22

)5)((2020 22 a

2/40 sma st 5.0

You are playing with a toy slider track.

The top of the “hump” is 1.2m above the level of the slider at the beginning of the track. The average force of friction between the 0.15kg slider & the track is 0.11N. The distance from point A to point B along the track is 2.3m.

You propel the slider by applying a constant force of 6.6N hoping to get it over the “hump”. How far must you push the slider to ensure that it makes it over the “hump”?

Top:

)2.1)(8.9)(15.0(gE

JEg 764.1

JEk 0

BUT….friction… Recall, work = energy

FdWE FricFric

)3.2)(11.0(FricW

JWE FricFric 253.0

JET 017.2

Bottom:

JET 017.2

Recall, energy= work

JWE appliedT 017.2

))(6.6(017.2 dmd 3056.0