Chapter 7
2. With speed v = 11200 m/s, we find
K
1 2 1
mv  (2.9 105 kg) (11200 m/s) 2  1.8 1013 J.
2
2
8. Using Eq. 7-8 (and Eq. 3-23), we find the work done by the water on the ice block:
W  F  d  (210 N) ˆi  (150 N) ˆj  (15 m) ˆi  (12 m)ˆj  (210 N) (15 m)  (150 N)(12 m)

 

3
 5.0 10 J.
9. By the work-kinetic energy theorem,
W  K 


1 2 1 2 1
mv f  mvi  (2.0 kg) (6.0 m/s) 2  (4.0 m/s) 2  20 J.
2
2
2


We note that the directions of v f and vi play no role in the calculation.
12. (a) From Eq. 7-6, F = W/x = 3.00 N (this is the slope of the graph).
(b) Equation 7-10 yields K = Ki + W = 3.00 J + 6.00 J = 9.00 J.
16. The change in kinetic energy can be written as
K 
1
1
m(v 2f  vi2 )  m(2a x)  ma x
2
2
v2f  vi2  2a x from Table 2-1. From the figure, we see that
 K  (0  30) J   30 J when x   5 m . The acceleration can then be obtained as
where we have used
a
K
( 30 J)

  0.75 m/s 2 .
m x (8.0 kg)(5.0 m)
The negative sign indicates that the mass is decelerating. From the figure, we also see
that when x  5 m the kinetic energy becomes zero, implying that the mass comes to rest
momentarily. Thus,
v02  v 2  2a x  0  2( 0.75 m/s 2 )(5.0 m)  7.5 m 2 /s 2 ,
or v0  2.7 m/s . The speed of the object when x = 3.0 m is
v  v02  2ax  7.5 m2 /s2  2( 0.75 m/s2 )(3.0 m)  12 m/s  3.5 m/s .
19. Equation 7-15 applies, but the wording of the problem suggests that it is only
necessary to examine the contribution from the rope (which would be the “Wa” term in
Eq. 7-15):
Wa = (50 N)(0.50 m) = 25 J
(the minus sign arises from the fact that the pull from the rope is anti-parallel to the
direction of motion of the block). Thus, the kinetic energy would have been 25 J greater
if the rope had not been attached (given the same displacement).
20. From the figure, one may write the kinetic energy (in units of J) as a function of x as
K  K s  20 x  40  20 x .
Since W  K  Fx  x , the component of the force along the force along +x is
Fx  dK / dx  20 N. The normal force on the block is FN  Fy , which is related to the
gravitational force by
mg  Fx2  ( Fy )2 .
(Note that FN points in the opposite direction of the component of the gravitational force.)
With an initial kinetic energy K s  40.0 J and v0  4.00 m/s , the mass of the block is
m
2Ks
2(40.0 J)

 5.00 kg.
2
v0
(4.00 m/s) 2
Thus, the normal force is
Fy  (mg )2  Fx2  (5.0 kg)2 (9.8 m/s 2 )2  (20 N) 2  44.7 N  45 N.
26. We make use of Eq. 7-25 and Eq. 7-28 since the block is stationary before and after
the displacement. The work done by the applied force can be written as
Wa  Ws 
1
k ( x 2f  xi2 ) .
2
The spring constant is k  (80 N) /(2.0 cm)=4.0 103 N/m. With Wa  4.0 J , and
xi   2.0 cm , we have
xf  
2Wa
2(4.0 J)
 xi2  
 ( 0.020 m) 2   0.049 m   4.9 cm.
3
k
(4.0 10 N/m)
30. Hooke’s law and the work done by a spring is discussed in the chapter. We apply the
work-kinetic energy theorem, in the form of K  Wa  Ws , to the points in Figure 7-35 at
x = 1.0 m and x = 2.0 m, respectively. The “applied” work Wa is that due to the constant
force P .
1
4 J  P(1.0 m)  k (1.0 m) 2
2
1
0  P(2.0 m)  k (2.0 m) 2 .
2
(a) Simultaneous solution leads to P = 8.0 N.
(b) Similarly, we find k = 8.0 N/m.
34. According to the graph the acceleration a varies linearly with the coordinate x. We
may write a = x, where  is the slope of the graph. Numerically,
20 m / s2

 2.5 s2 .
8.0 m
The force on the brick is in the positive x direction and, according to Newton’s second
law, its magnitude is given by F  ma  m x. If xf is the final coordinate, the work done
by the force is
m 2 (10 kg)(2.5 s 2 )
xf 
(8.0 m) 2  8.0 102 J.
0
0
2
2
36. From Eq. 7-32, we see that the “area” in the graph is equivalent to the work done.
Finding that area (in terms of rectangular [length  width] and triangular
[ 21 base  height] areas) we obtain
xf
xf
W   F dx  m  x dx 
W  W0 x2  W2 x4  W4 x6  W6 x8  (20 10  0  5) J  25 J.
 


45. The power associated with force F is given by P  F  v , where v is the velocity
of the object on which the force acts. Thus,
P  F  v  Fv cos   (122 N)(5.0 m/s)cos37  4.9 10 2 W.