Nguyen, Thanh – Homework 8 – Due: Oct 2 2007, 7:00 pm – Inst: D Weathers
1
This
printout
should
have
11
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Solve problems 8 and 9 using work and
energy concepts.
Assume in problem 9 that
the force applied to the car is constant.
001
(part 1 of 1) 10 points
A cheerleader lifts his 28
.
3 kg partner straight
up off the ground a distance of 0
.
27 m before
releasing her.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If he does this 31 times, how much work has
he done?
Correct answer: 2321
.
34 J.
Explanation:
The work done in lifting the cheerleader
once is
W
1
=
m g h
= (28
.
3 kg)(9
.
8 m
/
s
2
)(0
.
27 m)
= 74
.
8818 J
.
The work required to lift her
n
= 31 times is
W
=
n W
1
= (31)(74
.
8818 J) = 2321
.
34 J
.
keywords:
002
(part 1 of 1) 10 points
A shopper in a supermarket pushes a cart
with a force of 32
.
2 N directed at an angle of
26
.
7
◦
downward from the horizontal.
Find the work done by the shopper as she
moves down a 48
.
5 m length of aisle.
Correct answer: 1
.
39518 kJ.
Explanation:
Given :
F
shopper
= 32
.
2 N
,
θ
= 26
.
7
◦
,
and
d
= 48
.
5 m
.
The force is applied at an angle to the
displacement, so the net work is
W
shopper
=
F
shopper
d
cos
θ
= (32
.
2 N) (48
.
5 m) (cos 26
.
7
◦
)
1 kJ
1000 J
¶
=
1
.
39518 kJ
.
keywords:
003
(part 1 of 2) 10 points
A force
~
F
=
F
x
ˆ
ı
+
F
y
ˆ
acts on a particle that
undergoes a displacement of
~s
=
s
x
ˆ
ı
+
s
y
ˆ
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Weathers
 Energy, Force, Work, Correct Answer, θ, Thanh

Click to edit the document details