CK-12-Physics - Intermediate

http://www.ck12.org Chapter 6. Work and Energy

FIGURE 6.18

KE=

1

2

mv^2 =

1

2

( 0 .145kg)

(

15 .0m/s^2

) 2

= 16 .3 J

c. Use the Conservation of Mechanical Energy to find the maximum height the ball reaches.

Answer: Ei=Ef,

Ei=KEi+PEi=

1

2

mv^2 + 0

Ef=KEf+PEf= 0 +mgh

1

2

mv^2 =mgh,

1

2

v^2 =gh,h=

v^2

2 g

=

(15m/s)^2

( 2 )(10m/s^2 )

= 11 .2m

d. What is theKEof the ball at position 5.00 m?

Answer:ThePEof the ball at this position is:PE=mgh= ( 0 .145kg)(10m/s^2 )( 5 .00m) = 7 .2J

The total energy of the ball is 16.3 J and therefore theKEis: 16. 3 − 7. 1 = 9 .2 J

e. What is the speed of the ball at position 5.00 m?

Answer:

KE=

1

2

mv^2 ,

7. 11 =

1

2

( 0 .145kg)v^2

v= 9 .9 m/s