PH211 Chapter 6 Solutions
6.5
.
IDENTIFY: The gravity force is constant and the displacement is along a straight line, so W  Fs cos.
SET UP: The displacement is upward along the ladder and the gravity force is downward, so
  1800  300  1500. w  mg  735 N.
EXECUTE: (a) W  (735 N)(275 m)cos1500  - 1750 J.
(b) No, the gravity force is independent of the motion of the painter.
EVALUATE: Gravity is downward and the vertical component of the displacement is upward, so the gravity
force does negative work.
6.23
.
IDENTIFY: Apply W  Fs cos and Wtot  K .
SET UP:   0
EXECUTE: From Eqs. (6.1), (6.5) and (6.6), and solving for F,
2
2
2
2
1
1
K 2 m(v2  v1 ) 2 (800kg)((600m/s)  (400m/s) )


 320 N.
s
s
(250m)
EVALUATE: The force is in the direction of the displacement, so the force does positive work and the kinetic
energy of the object increases.
F
6.29.
IDENTIFY: Wtot  K2  K1. Only friction does work.
SET UP: Wtot  W fk  - k mgs. K2  0 (car stops). K1  12 mv02 .
EXECUTE: (a) Wtot  K2  K1 gives k mgs  - 12 mv02. s 
(b) (i) kb  2ka . sk 
v02
2k g
.
 
v02
 constant so sa ka  sb kb . sb   ka  sa  sa /2. The minimum stopping
2g
 kb 
2
v 
. sb  sa  0b   4sa . The
2
2
2
v0 2k g
v0a v0b
 v0a 
s
1
 constant, so
stopping distance would become 4 times as great. (iii) v0b  2v0a , kb  2ka . 2k 
2g
v0
distance would be halved. (ii) v0b  2v0a .
s

1
 constant, so
sa

sb
2
   v 
1 2
. sb  sa  ka  0b   sa    2   2sa . The stopping distance would double.
2
2
v0a
v0b
2
 kb  v0a 
EVALUATE: The stopping distance is directly proportional to the square of the initial speed and indirectly
proportional to the coefficient of kinetic friction.
sa ka
6.43 .

sb kb
IDENTIFY and SET UP: Apply Eq. (6.6). Let point 1 be where the sled is released and point 2 be at x  0 for
part (a) and at x  0200 m for part (b). Use Eq. (6.10) for the work done by the spring and calculate K 2  Then
K2  12 mv22 gives v2 
EXECUTE: (a) Wtot  K2  K1 so K2  K1  Wtot
K1  0 (released with no initial velocity), K2  12 mv22
The only force doing work is the spring force. Eq. (6.10) gives the work done on the spring to move its end from
x1 to x2  The force the spring exerts on an object attached to it is F  - kx, so the work the spring does is
Wspr  -
 12 kx22  12 kx12   12 kx12  12 kx22. Here x  - 0375 m and x
1
2
 0. Thus
Wspr  12 (4000 N/m)(0375 m)2  0  281 J.
K2  K1  Wtot  0  281 J  281 J
Then K2  12 mv22 implies v2 
2K2
2(281 J)

 283 m/s.
m
700 kg
(b) K2  K1  Wtot
K1  0
Wtot  Wspr  12 kx12  12 kx22. Now x2  0200 m, so
Wspr  12 (4000 N/m)(0375 m)2  12 (4000 N/m)(0200 m) 2  281 J  80 J  201 J
Thus K2  0  201 J  201 J and K2  12 mv22 gives v2 
2K2
2(201 J)

 240 m/s
m
700 kg
EVALUATE: The spring does positive work and the sled gains speed as it returns to x  0 More work is done
during the larger displacement in part (a), so the speed there is larger than in part (b).
6.45.
IDENTIFY and SET UP: Apply Eq. (6.6) to the glider. Work is done by the spring and by gravity. Take point 1
to be where the glider is released. In part (a) point 2 is where the glider has traveled 1.80 m and K2  0 There
are two points shown in Figure 6.45a. In part (b) point 2 is where the glider has traveled 0.80 m.
EXECUTE: (a) Wtot  K2  K1  0. Solve for x1, the amount the spring is initially compressed.
Wtot  Wspr  Ww  0
So Wspr  2 Ww
(The spring does positive work on the glider since
the spring force is directed up the incline, the same
as the direction of the displacement.)
Figure 6.45a
The directions of the displacement and of the gravity force are shown in Figure 6.45b.
Ww  (wcos )s  (mg cos1300) s
Ww  (00900 kg)(980 m/s2 )(cos1300)(180 m)  - 1020 J
(The component of w parallel to the incline is
directed down the incline, opposite to the
displacement, so gravity does negative work.)
Figure 6.45b
Wspr  - Ww  + 1020 J
2(1020 J)

 00565 m
k
640 N/m
(b) The spring was compressed only 0.0565 m so at this point in the motion the glider is no longer in contact
with the spring. Points 1 and 2 are shown in Figure 6.45c.
Wspr  12 kx12 so x1 
2Wspr
Wtot  K2  K1
K2  K1  Wtot
K1  0
Figure 6.45c
Wtot  Wspr  Ww
From part (a), Wspr  1020 J and
Ww  (mg cos1300)s  (00900 kg)(980 m/s2 )(cos1300)(080 m)  0454 J
Then K2  Wspr  Ww  + 1020 J  0454 J  + 057 J
EVALUATE: The kinetic energy in part (b) is positive, as it must be. In part (a), x2  0 since the spring force is
no longer applied past this point. In computing the work done by gravity we use the full 0.80 m the glider
moves.
6.50.
IDENTIFY: Knowing the rate at which energy is consumed, we want to find out the total energy used.
SET UP: Find the elapsed time t in each case by dividing the distance by the speed, t  d/v. Then calculate
the energy as W  Pt
EXECUTE: Running: t  (50 km)/(10 km/h)  050 h  18  103 s The energy used is
W  (700 W)(18  103 s)  13  106 J
50 km  3600 s 
3
3
6

  60  10 s The energy used is W  (290 W)(60  10 s)  17  10 J
30 km/h  h 
EVALUATE: The less intense exercise lasts longer and therefore burns up more energy than the intense
exercise.
Walking: t 
6.57.
IDENTIFY: To lift the skiers, the rope must do positive work to counteract the negative work developed by the
component of the gravitational force acting on the total number of skiers,
Frope  Nmg sin  .
SET UP: P  F||v  Fropev
EXECUTE: Prope  Fropev  [ Nmg (cos )]v.

 1 m/s  
4
Prope  [(50 riders)(700 kg)(980 m/s2 )(cos750)] (120 km/h) 
  . Prope  296 10 W  296 kW.
 360 km/h  

EVALUATE: Some additional power would be needed to give the riders kinetic energy as they are accelerated
from rest.
6.84.
IDENTIFY: Apply Wtot  K2  K1. W  Fs cos.
SET UP: The students do positive work, and the force that they exert makes an angle of 300 with the
direction of motion. Gravity does negative work, and is at an angle of 1200 with the chair’s motion.
EXECUTE: The total work done is
Wtot  ((600 N) cos300  (850 kg)(980 m/s2 ) cos1200)(250 m)  2578 J, and so the speed at the top of the
ramp is v2  v12 
2Wtot
2(2578 J)
 (200 m/s)2 
 317 m/s
m
(850 kg)
6.86.
EVALUATE: The component of gravity down the incline is mg sin30  417 N and the component of the push
up the incline is (600 N)cos30  520 N. The force component up the incline is greater than the force
component down the incline; the net work done is positive and the speed increases.
IDENTIFY: Apply Wtot  K2  K1 to the system of the two blocks. The total work done is the sum of that done
by gravity (on the hanging block) and that done by friction (on the block on the table).
SET UP: Let h be the distance the 6.00 kg block descends. The work done by gravity is (6.00 kg)gh and the
work done by friction is k (800 kg) gh.
EXECUTE: Wtot  (600 kg  (025)(8 00 kg))(9 80 m/s2 )(1 50 m)  58 8 J This work increases the kinetic
1
2(588 J)
energy of both blocks: Wtot  (m1  m2 )v 2 , so v 
 290 m/s
2
(1400 kg)
EVALUATE: Since the two blocks are connected by the rope, they move the same distance h and have the same
speed v.
6.95.
IDENTIFY and SET UP: For part (a) calculate m from the volume of blood pumped by the heart in one day. For
part (b) use W calculated in part (a) in Eq. (6.15).
EXECUTE: (a) W  mgh, as in Example 6.10. We need the mass of blood lifted; we are given the volume
 1 103 m3 
3
V  (7500 L) 
  750 m 

1
L


m  density  volume  (105 103 kg/m3 )(750 m3 )  7875 103 kg
Then W  mgh  (7875 103 kg)(980 m/s2 )(163 m)  126 105 J
W
126  105 J

 146 W
t
(24 h)(3600 s/h)
EVALUATE: Compared to light bulbs or common electrical devices, the power output of the heart is rather
small.
(b) Pav 