Download HW5

PHYS-207 Honors Section-LL HW5 Solutions
7.4. We apply the equation x (t )  x0  v0t  12 at , found in Table 2-1. Since at t = 0 s, x0 = 0, and
2
v0  12 m/s , the equation becomes (in unit of meters)
x(t )  12t  12 at 2 .
x  10 m when t  1.0 s , the acceleration is found to be a   4.0 m/s 2 . The fact that
a  0 implies that the bead is decelerating. Thus, the position is described by x(t )  12t  2.0t 2 .
With
Differentiating x with respect to t then yields
v(t ) 
dx
 12  4.0t .
dt
Indeed at t =3.0 s, v(t  3.0)  0 and the bead stops momentarily. The speed at
t  10 s is
v(t  10)   28 m/s , and the corresponding kinetic energy is
1
1
K  mv 2  (1.8 10 2 kg)(  28 m/s) 2  7.1 J.
2
2
 

7.15. (a) The forces are constant, so the work done by any one of them is given by W  F  d , where d is
the displacement. Force

F1 is in the direction of the displacement, so
W1  F1d cos 1  (5.00 N)(3.00m)cos0  15.0 J.

F2 makes an angle of 120° with the displacement, so
W2  F2d cos 2  (9.00 N)(3.00m)cos120  13.5 J.

Force F3 is perpendicular to the displacement, so
Force
W3 = F3d cos 3 = 0 since cos 90° = 0.
The net work done by the three forces is
W  W1  W2  W3  15.0 J  13.5 J  0  1.50 J.
(b) If no other forces do work on the box, its kinetic energy increases by 1.50 J during the displacement.

7.17. We use F to denote the upward force exerted by the cable on the astronaut. The force of the cable
is upward and the force of gravity is mg downward. Furthermore, the acceleration of the astronaut is a =
g/10 upward. According to Newton’s second law, the force is given by
F  mg  ma  F  m( g  a ) 
11
mg ,
10
in the same direction as the displacement. On the other hand, the force of gravity has magnitude
Fg  mg and is opposite in direction to the displacement.


(a) Since the force of the cable F and the displacement d are in the same direction, work done by
WF  Fd 

F is
11mgd 11 (72 kg)(9.8 m/s 2 )(15 m)

 1.164 104 J  1.2 104 J .
10
10
(b) Using Eq. 7-7, the work done by gravity is
Wg  Fg d  mgd   (72 kg)(9.8 m/s2 )(15 m)  1.058 104 J  1.1104 J
(c) The total work done is the sum of the two works:
Wnet  WF  Wg  1.164 104 J 1.058 104 J  1.06 103 J  1.1103 J .
Since the astronaut started from rest, work-kinetic energy theorem says that it’s the final kinetic energy.
(d) Since K 
1
2
mv 2 , her final speed is v 
2K
2(1.06 103 J)

 5.4 m/s .
m
72 kg
Note: For a general upward acceleration a, the net work done is
Wnet  WF  Wg  Fd  Fg d  m( g  a )d  mgd  mad .
Since Wnet   K  mv / 2, by the work-kinetic energy theorem, the speed of the astronaut would be
2
v  2ad , which is independent of the mass of the astronaut.
7.31. (a) As the body moves along the x axis from xi = 3.0 m to xf = 4.0 m the work done by the force is
xf
xf
xi
xi
W   Fx dx   6 x dx  3( x 2f  xi2 )  3 (4.02  3.02 )  21 J.
According to the work-kinetic energy theorem, this gives the change in the kinetic energy:
W  K 
1
m v 2f  vi2
2
d
i
where vi is the initial velocity (at xi) and vf is the final velocity (at xf). The theorem yields
vf 
2W 2
2(21 J)
 vi 
 (8.0 m/s)2  6.6 m/s.
m
2.0 kg
(b) The velocity of the particle is vf = 5.0 m/s when it is at x = xf. The work-kinetic energy theorem is used

to solve for xf. The net work done on the particle is W  3 x f  xi
d
i
3 x 2f  xi2 
xf  
2
2
 , so the theorem leads to
1
m v 2f  vi2 .
2
d
i
m 2 2
2.0 kg
v f  vi   xi2  
(5.0 m/s)2  (8.0 m/s)2   (3.0 m)2  4.7 m.


6
6 N/m
7.46. Recognizing that the force in the cable must equal the total weight (since there is no acceleration),
P  Fv cos   mg
where we have
  0
x I
F
G
Ht J
K
(both the force of the cable and the elevator’s motion are upward). Thus,
 210 m 
5
P  (3.0 103 kg)(9.8 m/s 2 ) 
  2.7 10 W.
 23 s 
7.63. The forces involved are: gravitational force on the crate, normal force on the crate, and the force

applied by the worker. The work done by a force F on an object as it moves through a displacement
r
r r

W  F  d  Fd cos , where  is the angle between F and d .
(a) The applied force is parallel to the incline. Thus, using Eq. 7-6, the work done on the crate by the
worker’s applied force is
Wa  Fd cos 0   (209 N)(1.50 m)  314 J.
(b) Using Eq. 7-12, we find the work done by the gravitational force to be
Wg  Fg d cos(90   25 )  mgd cos115 
 (25.0 kg)(9.8 m/s 2 )(1.50 m) cos115 
  155 J.
r
d is
(c) The angle between the normal force and the direction of motion is 90º, so the work it does is zero:
WN  FN d cos90   0 .
(d) The total work done on the crate is the sum of all three works:
W  Wa  Wg  WN  314 J  ( 155 J)  0 J  158 J .
Note: By the work-kinetic energy theorem, if the crate is initially at rest, then its kinetic energy after
having moved 1.50 m up the incline is K  W  158 J , and the speed of the crate at that instant is
v  2K / m  2(158 J) / 25.0 kg  3.56 m/s .
8.9. We use Eq. 8-17, representing the conservation of mechanical energy (which neglects friction and
other dissipative effects).
(a) In Problem 9-2, we found UA = mgh (with the reference position at C). Referring again to Fig. 8-27, we
see that this is the same as U0, which implies that KA = K0 and thus that
vA = v0 = 17.0 m/s.
(b) In the solution to Problem 9-2, we also found U B  mgh 2. In this case, we have
K0  U 0  K B  U B
F
IJ
G
HK
1 2
1
h
mv0  mgh  mv B2  mg
2
2
2
vB  v02  gh  (17.0 m/s)2  (9.80 m/s 2 )(42.0 m)  26.5 m/s.
(c) Similarly,
vC  v02  2 gh  (17.0 m/s)2  2(9.80 m/s 2 )(42.0 m)  33.4 m/s.
(d) To find the “final” height, we set Kf = 0. In this case, we have
K0  U 0  K f  U f
1 2
mv0  mgh  0  mgh f
2
v02
(17.0 m/s)2
which yields h f  h 
 42.0 m 
 56.7 m.
2g
2(9.80 m/s 2 )
(e) It is evident that the above results do not depend on mass. Thus, a different mass for the coaster must
lead to the same results.
8.31. The reference point for the gravitational potential energy Ug (and height h) is at the block when the
spring is maximally compressed. When the block is moving to its highest point, it is first accelerated by
the spring; later, it separates from the spring and finally reaches a point where its speed vf is
(momentarily) zero. The x axis is along the incline, pointing uphill (so x0 for the initial compression is
negative); its origin is at the relaxed position. We use SI units, so k = 1960 N/m and x0 = –0.200 m.
(a) The elastic potential energy is
1
2
kx02  39.2 J .
(b) Since initially Ug = 0, the change in Ug is the same as its final value mgh where m = 2.00 kg. That this
must equal the result in part (a) is made clear in the steps shown in the next part. Thus, Ug = Ug = 39.2 J.
K0  U 0  K f  U f
(c) The principle of mechanical energy conservation leads to
1
0  kx02  0  mgh
2
which yields h = 2.00 m. The problem asks for the distance along the incline, so we have d = h/sin 30° =
4.00 m.