Problem Solution: A one kilogram object, initially at rest, falls from a height 180 meters above the earth.

Approximating the acceleration of gravity as g =10 m/s2, the object will strike the earth in 6 seconds.

(a)  Compute the instantaneous height h and the speed v at one second intervals over the physical trajectory.

From this compute the instantaneous values for the kinetic and potential energies, and their sum and difference.

Total the differences for the seven intervals

t

v=gt

h=180-gt2/2

KE=mv2/2

PE=mgh

KE+PE

KE-PE

0

0

180

0

1800

1800

-1800

1

10

175

50

1750

1800

-1700

2

20

160

200

1600

1800

-1400

3

30

135

450

1350

1800

-900

4

40

100

800

1000

1800

-200

5

50

55

1250

550

1800

+700

6

60

0

1800

0

1800

+1800

# Total

-3500

(b)  Repeat the calculation using a non-physical path in which the object travels the distance 180 meters in

6 seconds at the constant speed v=30 m/s.

t

v=30

h=180-vt

KE=mv2/2

PE=mgh

KE+PE

KE-PE

0

30

180

450

1800

2250

-1350

1

30

150

450

1500

1950

-1050

2

30

120

450

1200

1650

-750

3

30

90

450

900

1350

-450

4

30

60

450

600

1050

-150

5

30

30

450

300

750

+150

6

30

0

450

0

450

+450

# Total

-3150

(c)  Determine speed of fall as an object thrown up at 5 m/s, which falls with an acceleration of

a = (7/6)g = 70/6 m/s2.  Thus

v = -5 + (70/6)t                        h = 180 + 5t – (35/6)t2

 t v h KE=mv2/2 PE=mgh KE+PE KE-PE 0 -5 180 12.5 1800.0 1812 -1788 1 6.67 179.17 22.2 1791.7 1814 -1769 2 18.33 166.67 168.1 1666.7 1835 -1499 3 30 142.5 450.0 1425.0 1875 -975 4 41.67 106.67 868.1 1066.7 1935 -199 5 53.33 59.17 1422.2 591.7 2014 +830 6 65 0 2112.5 0 2113 +2113 Total -3286