This preview shows pages 1–2. Sign up to view the full content.
1
Chapter 6 Conservation of Energy
2. A 20.0 N force is applied at an angle of 40.0 degrees above the horizontal to a 4.00 kg box. The box moves a
horizontal distance of 4.00 meters. The work done by the 20.0 N force is:
A. 75.0 J.
B. 61.3 J.
C. 50.1 J.
D. 46.3 J.
E. 40.5 J
J
3
.
61
40
cos
)
4
)(
20
(
cos
m
N
d
F
d
F
W
3. A 30.00 kg box slides up a 12.00 meters long incline at an angle of 30.00 degrees above the horizontal. The
change in gravitational potential energy is:
A. 2110 J.
B. 1467 J.
C. 1080 J.
D. 1764 J.
E. 760.0 J.
We need the vertical height h, which is h = d sin
= (12m) sin 30
o
= 6m. Then
E
pot
= m g h = (30 kg)(9.8 m s
2
)(6 m)=1,764 J
5. A horizontal force is applied to a 4.0 kg box. The box starts from rest moves a horizontal distance of 8.0
meters and obtains a velocity of 6.0 m/s. The system is frictionless. The horizontal force is:
A. 3.0 N.
B. 5.0 N.
C. 7.0 N.
D. 9.0 N.
E. 11 N.
This can be done in two ways:
1)
Using kinematics and Newton
’
s 2
nd
law: First find the acceleration. Using
v
f
2
v
i
2
= 2
a
x, we find
a
= v
f
2
/2
x = (6 m/s)
2
/(2*8m) = 2.25 m/s
2
in this case.
Then the force is given by F = m a = (4.0 kg)(2.25 m/s
2
) = 9.0 N.
2)
Using energy considerations. Since we have no friction in this case, the mechanical energy is conserved. Therefore the work done on the
system is all converted into kinetic energy. This means W = E
kin
. The kinetic energy is E
kin
= ½ m v
2
= 72 J. The work is W = F d and
therefore F = W/d = E
kin
/d = (72 J)/(8.0 m) = 9.0N
10. As a hiker descends a hill, the work done by gravity on the hiker is:
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 '09
 ZHIXIANZHOU,
 Conservation Of Energy, Energy, Force, Work

Click to edit the document details