pasha (sep635) – HW 62 – Antoniewicz – (56445)
1
This printout should have 19 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001 (part 1 of 4) 2.0 points
An object at location
h
20
,
0
,
0
i
mmovesto
location
h
30
,
0
,
0
i
m. While it is moving it
is acted on by a constant Force oF
h
22
,
0
,
0
i
N.
How much work is done on the object by this
Force?
Correct answer:

220 J.
Explanation:
W
=
~
F
·
Δ
~
r
=
F
x
Δ
x
+
F
y
Δ
y
+
F
z
Δ
z
=(22
N
)(

30

(

20))m
=

220
002 (part 2 of 4) 1.0 points
Does the kinetic energy oF the object increase
or decrease?
1.
Decrease
correct
2.
Increase
3.
There is no change in the kinetic energy
oF the ball.
Explanation:
Consider the update Form oF the Conserva
tion oF Energy:
E
sys,f
=
E
sys,i
+
W
surr
.Th
e
system is the object, and the surroundings
are doing work on the object. The calculated
work due to the Force on the object is neg
ative. This means that the initial energy oF
the system is greater than the fnal energy oF
the system, i.e. the energy oF the system de
creased. The Force acts on the object in a
direction that opposes the initial motion oF
the object. Thus, the kinetic energy oF the
object decreases.
003 (part 3 of 4) 2.0 points
Now,
consider a di
f
erent object as it
moves From location
h
30
,
0
,
0
i
mtolocat
ion
h
20
,
0
,
0
i
m. While it is moving, it is acted
on by a constant Force oF
h
22
,
0
,
0
i
N. How
much work is done on the second object by
this Force?
Correct answer: 220 J.
Explanation:
W
=
~
F
•
Δ
~
r
=
F
x
Δ
x
+
F
y
Δ
y
+
F
z
Δ
z
=(22
N
)(

20

(

30))m
=220
004 (part 4 of 4) 1.0 points
Does the kinetic energy oF the second object
increase or decrease?
1.
There is no change in the kinetic energy
oF the ball.
2.
Increase
correct
3.
Decrease
Explanation:
Apply the update Form oF the Conservation
oF Energy once again. The work calculated is
positive, so the fnal energy state will have a
greater value than the initial energy state.
005 (part 1 of 4) 2.0 points
AballoFmass0
.
9kgFallsdownward. Initially
,
you observe it to be 4
.
75 m above the ground.
AFter a short time, the ball is just about to hit
the ground. During this interval, how much
work was done on the ball by the gravitational
Force? Use 9
.
8
m
s
2
as the acceleration due to
gravity.
Correct answer: 41
.
895 J.
Explanation:
W
=
~
F
·
Δ
~
r
=
F