Download CHAPTER 7 SOLUTION FOR PROBLEM 17 (a) Let F be the

CHAPTER 7
SOLUTION FOR PROBLEM 17
(a) Let F be the magnitude of the force exerted by the cable on the astronaut. The force of
the cable is upward and the force of gravity is mg is downward. Furthermore, the acceleration
of the astronaut is g/10, upward. According to Newton’s second law, F − mg = mg/10, so
F = 11mg/10. Since the force Fn and the displacement dn are in the same direction the work
done by Fn is
2
11mgd 11(72 kg)(9.8 m/s )(15 m)
=
= 1.16 × 104 J .
WF = F d =
10
10
(b) The force of gravity has magnitude mg and is opposite in direction to the displacement. Since
cos 180◦ = −1, it does work
2
Wg = −mgd = −(72 kg)(9.8 m/s )(15 m) = −1.06 × 104 J .
(c) The total work done is W = 1.16 × 104 J − 1.06 × 104 J = 1.1 × 103 J. Since the astronaut
started from rest the work-kinetic energy theorem tells us that this must be her final kinetic energy.
(d) Since K = 12 mv 2 her final speed is
v=
5
2K
=
m
2(1.1 × 103 J)
= 5.3 m/s .
72 kg
CHAPTER 7
SOLUTION FOR PROBLEM 24
(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
exf
8 xf
8 xf
e
2e
Fx dx =
−6x dx = −3x e = −3(x2f − x2i )
W =
xi
J
xi
2
= −3 (4.0) − (3.0)
2
o
xi
= −21 J .
According to the work-kinetic energy theorem, this is the change in the kinetic energy:
W = ∆K = 12 m(vf2 − vi2 ) ,
where vi is the initial velocity (at xi ) and vf is the final velocity (at xf ). The theorem yields
5
2W
2(−21 J)
+ vi2 =
+ (8.0 m/s)2 = 6.6 m/s .
vf =
m
2.0 kg
(b) The velocity of the particle is vf = 5.0 m/s when it is at x = xf . Solve the work-kinetic energy
theorem for xf . The net work done on the particle is W = −3(x2f − x2i ), so the work-kinetic
energy theorem yields −3(x2f − x2i ) = 12 m(vf2 − vi2 ). Thus
5
o
m 2
2.0 kg J
(5.0 m/s)2 − (8.0 m/s)2 + (3.0 m)2 = 4.7 m .
xf = − (vf − vi2 ) + x2i = −
6
6 N/m
CHAPTER 7
SOLUTION FOR PROBLEM 27
(a) The graph shows F as a function of x if x0 is positive.
The work is negative as the object moves from x = 0 to
x = x0 and positive as it moves from x = x0 to x = 2x0 .
Since the area of a triangle is 12 (base)(altitude), the work
done from x = 0 to x = x0 is − 12 (x0 )(F0 ) and the work done
from x = x0 to x = 2x0 is 12 (2x0 − x0 )(F0 ) = 12 (x0 )(F0 ). The
total work is the sum, which is zero.
(b) The integral for the work is
W =
8
2x0
0
F0
w
F (x)
F0
0
−F0
....
.....
.....
.....
.....
.
.
.
.
....
.....
.....
.....
.....
.
.
.
.
.
.....
....
.....
.....
....
.
.
.
.
....
.....
.....
.....
.....
.
.
.
.
.....
.....
W
W e2x0
w 2
e
x
x
− 1 dx = F0
− x ee
= 0.
x0
2x0
0
x0
2x0 x
CHAPTER 7
HINT FOR PROBLEM 11
The magnitude of the force is given by Newton’s second law: F = ma, where m is the mass of
the luge and rider and a is the magnitude of their acceleration. Since the force is constant and
directed oppositely to the displacement, the work it does is W = −F d, where d is the distance
traveled while stopping. According to the work-kinetic energy theorem this must be the change
in kinetic energy of the luge and rider and since the luge stops, −12 mv 2 = −F d, where v is the
initial speed of the luge. Solve for d. Use W = −12 mv 2 or W = −F d to calculate the work done
by the force.
J
ans: (a) 1.7 × 10o2 N; (b) 3.4 × 102 m; (c) −5.8 × 104 J; (d) 3.4 × 102 N; (e) 1.7 × 102 N;
(f) −5.8 × 104 J
CHAPTER 7
HINT FOR PROBLEM 21
The work done by the cable is given by W = T d, where T is the tension force of the cable and d
is the distance the elevator cab travels (d1 in part (a) and d2 in part (b)). According to Newton’s
second law the acceleration of the cheese (and also of the elevator) is a = FN /mc , where mc is
the mass of the cheese, and the tension force of the cable is T = me a, where me is the mass of
the elevator cab. (Strictly, it should be the mass of the cab and cheese together, but the mass of
the cheese is so small it may be neglected here.)
(a) Put d equal to d1 (= 2.40 m) and FN = 3.00 N, then solve for W .
(b) Put d = d2 (= 10.5 m) and W = 92.61 × 103 J, then solve for FN .
J
ans: (a) 25.9 kJ; (b) 2.45 N
o