kaplan (hmk378) – Homework 08b – Weathers – (17104)
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A Force
v
F
=
F
x
ˆ
ı
+
F
y
ˆ
acts on a particle that
undergoes a displacement oF
vs
=
s
x
ˆ
ı
+
s
y
ˆ
where
F
x
= 2 N,
F
y
=
−
3 N,
s
x
= 6 m, and
s
y
= 3 m.
±ind the work done by the Force on the
particle.
Correct answer: 3 J.
Explanation:
The work is given by
W
=
v
F
·
=
F
x
s
x
+
F
y
s
y
= (2 N) (6 m) + (
−
3 N) (3 m)
=
3 J
.
002
(part 2 oF 2) 10.0 points
±ind the angle between
v
F
and
.
Correct answer: 82
.
875
◦
.
Explanation:
Since
W
=
v
F
·
=
v
v
v
v
F
v
v
v


cos
θ
θ
= cos
−
1
W
v
v
v
v
F
v
v
v


.
v
v
v
v
F
v
v
v
=
r
F
2
x
+
F
2
y
=
r
(2 N)
2
+ (
−
3 N)
2
= 3
.
60555 N
and


=
r
s
2
x
+
s
2
y
=
R
6 m
2
+ 3 m
2
= 6
.
7082 m
,
so
θ
= cos
−
1
b
3 J
(3
.
60555 N) (6
.
7082 m)
B
=
82
.
875
◦
.
003
10.0 points
A shopper in a supermarket pushes a cart
with a Force oF 32
.
3 N directed at an angle oF
23
.
7
◦
downward From the horizontal.
±ind the work done by the shopper as she
moves down a 51
.
6 m length oF aisle.
Correct answer: 1
.
52612 kJ.
Explanation:
Given :
F
shopper
= 32
.
3 N
,
θ
= 23
.
7
◦
,
and
d
= 51
.
6 m
.
The Force is applied at an angle to the
displacement, so the net work is
W
shopper
=
F
shopper
d
cos
θ
= (32
.
3 N) (51
.
6 m) (cos 23
.
7
◦
)
p
1 kJ
1000 J
P
=
1
.
52612 kJ
.
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.
 Fall '08
 Weathers
 mechanics, Energy, Force, Work, Correct Answer, θ

Click to edit the document details