This document is a cache from http://iweb.tntech.edu/murdock/books/v1chap6.pdf

# Work, Kinetic Energy and Potential Energy

Document source : iweb.tntech.edu

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 All Pages

6.2. WORKED EXAMPLES
141
v
0
=
20
m/s
v
34
o
v
y
=0
(Max height)
h
Figure 6.7:
Snowball is launched at angle of 34
in Example 11.
This occurs every second, so gravity does work at a rate of
P
grav
=
mgh
t
=
5.88 × 10
8
J
1 s
= 5.88 × 10
8
W
As we see later, this is also the rate at which the water loses potential energy. This energy
can be converted to other forms, such as the electrical energy to make a light bulb function.
In this highly idealistic example, all of the energy is converted to electrical energy.
A 60 W light bulb uses energy at a rate of 60
J
s
= 60 W. We see that Niagara Falls
puts out energy at a rate much bigger than this! Assuming all of it goes to the bulbs, then
dividing the total energy consumption rate by the rate for one bulb tells us that
N =
5.88 × 10
8
W
60 W
= 9.8 × 10
6
bulbs can be lit.
6.2.6
Conservation of Mechanical Energy
11. A 1.50 kg snowball is shot upward at an angle of 34.0
to the horizontal with
an initial speed of 20.0
m
s
. (a) What is its initial kinetic energy? (b) By how much
does the gravitational potential energy of the snowball­Earth system change as
the snowball moves from the launch point to the point of maximum height? (c)
What is that maximum height?
[HRW5 8-31]
(a) Since the initial speed of the snowball is 20.0
m
s
, we have its initial kinetic energy:
K
i
=
1
2
mv
2
0
=
1
2
(1.50 kg)(20.0
m
s
)
2
= 300. J
(b) We need to remember that since this projectile was not fired straight up, it will still
have some kinetic energy when it gets to maximum height! That means we have to think a
little harder before applying energy principles to answer this question.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 All Pages

Summary :

## This occurs every second, so gravity does work at a rate of P grav = mgh t = 5.88 × 10 8 J 1 s = 5.88 × 10 8 W As we see later, this is also the rate at which the water loses potential energy. Assuming all of it goes to the bulbs, then dividing the total energy consumption rate by the rate for one bulb tells us that N = 5.88 × 10 8 W 60 W = 9.8 × 10 6 bulbs can be lit.

Tags : rate,snowball,initial,height,588,kinetic,hae,bulb,maximum,does,since,example,much
Related Documents

 Terms    |    Link pdf-search-files.com    |    Site Map    |    Contact    All books are the property of their respective owners. Please respect the publisher and the author for their creations if their books copyrighted © 2009 pdf-search-files.com