Chapter 7

7
POTENTIAL ENERGY AND ENERGY CONSERVATION
7.1.
IDENTIFY: U grav = mgy so ΔU grav = mg( y2 − y1 )
SET UP:
+ y is upward.
EXECUTE:
(a) ΔU = (75 kg)(9.80 m/s 2 )(2400 m − 1500 m) = +6.6 × 105 J
(b) ΔU = (75 kg)(9.80 m/s 2 )(1350 m − 2400 m) = −7.7 × 105 J
EVALUATE:
U grav increases when the altitude of the object increases.
7.5.
IDENTIFY and SET UP:
Use energy methods.
(a) K1 + U1 + Wother = K 2 + U 2 . Solve for K 2 and then use K 2 = 12 mv22 to obtain v2 .
Wother = 0 (The only force on the
ball while it is in the air is gravity.)
K1 = 12 mv12 ; K 2 = 12 mv22
U1 = mgy1 , y1 = 22.0 m
U 2 = mgy2 = 0, since y2 = 0
for our choice of coordinates.
Figure 7.5
EXECUTE:
1
2
mv12 + mgy1 = 12 mv22
v2 = v12 + 2gy1 = (12.0 m/s)2 + 2(9.80 m/s 2 )(22.0 m) = 24.0 m/s
EVALUATE:
The projection angle of 53.1° doesn’t enter into the calculation. The kinetic energy
depends only on the magnitude of the velocity; it is independent of the direction of the velocity.
(b) Nothing changes in the calculation. The expression derived in part (a) for v2 is independent of the
angle, so v2 = 24.0 m/s, the same as in part (a).
(c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect.
7.11.
IDENTIFY:
Apply Eq.(7.7) to the motion of the car.
SET UP:
Take y = 0 at point A. Let point 1 be A and point 2 be B.
K1 + U1 + Wother = K 2 + U 2
EXECUTE:
U1 = 0, U 2 = mg(2R) = 28,224 J, Wother = W f
K1 = 12 mv12 = 37,500 J, K 2 = 12 mv22 = 3840 J
The work-energy relation then gives W f = K 2 + U 2 − K1 = −5400 J.
EVALUATE:
Friction does negative work. The final mechanical energy (K 2 + U 2 = 32,064 J) is less
than the initial mechanical energy (K1 + U1 = 37,500 J) because of the energy removed by friction
work.
7.14.
IDENTIFY:
Only gravity does work, so apply Eq.(7.4). Use


∑ F = ma to calculate the tension.
SET UP:
Let y = 0 at the bottom of the arc. Let point 1 be when the string makes a 45∞angle with the
vertical and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial
acceleration arad = v 2 / r
EXECUTE:
(a) At the top of the swing, when the kinetic energy is zero, the potential energy (with
respect to the bottom of the circular arc) is mgl(1 − cos ), where l is the length of the string and  is
the angle the string makes with the vertical. At the bottom of the swing, this potential energy has
become kinetic energy, so mgl(1 − cos) = 12 mv 2 , or
v = 2gl(1 − cos) = 2(9.80 m/s 2 ) (0.80 m) (1 − cos 45°) = 2.1 m/s .
(b) At 45° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to
the radial component of the weight, or mg cosθ = (0.12 kg) (9.80 m/s 2 ) cos 45° = 0.83 N.
(c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial
acceleration,
mg + mv22 l = mg(1 + 2(1 − cos 45°)) = 1.9 N
EVALUATE:
When the string passes through the vertical, the tension is greater than the weight
because the acceleration is upward.
7.16.
IDENTIFY:
Use the information given in the problem with F = kx to find k. Then U el = 12 kx 2 .
SET UP:
x is the amount the spring is stretched. When the weight is hung from the spring, F = mg .
EXECUTE:
k =
F mg (3.15 kg)(9.80 m/s 2 )
=
=
= 2205 N/m .
x
x
0.1340 m − 0.1200 m
2U el
2(10.0 J)
=±
= ±0.0952 m = ±9.52 cm . The spring could be either stretched 9.52 cm
k
2205 N/m
or compressed 9.52 cm. If it were stretched, the total length of the spring would be
12.00 cm + 9.52 cm = 21.52 cm . If it were compressed, the total length of the spring would be
12.00 cm − 9.52 cm = 2.48 cm .
EVALUATE:
To stretch or compress the spring 9.52 cm requires a force F = kx = 210 N .
x=±
7.19.
IDENTIFY and SET UP:
Use energy methods. There are changes in both elastic and gravitational
potential energy; elastic; U = 12 kx 2 , gravitational: U = mgy.
2U
2(3.20 J)
=
= 0.0632 m = 6.32 cm
k
1600 N/m
(b) Points 1 and 2 in the motion are sketched in Figure 7.19.
K1 + U1 + Wother = K 2 + U 2
EXECUTE:
(a) U = 12 kx 2 so x =
Wother = 0 (Only work is that done
by gravity and spring force)
K1 = 0, K 2 = 0
y = 0 at final position of book
(
)
U1 = mg h + d , U 2 = 12 kd 2
Figure 7.19
0 + mg(h + d) + 0 = 12 kd 2
The original gravitational potential energy of the system is converted into potential energy of the
compressed spring.
1
2
kd 2 − mgd − mgh = 0
d=

1 
1
2
 mg ± (mg) + 4  k  (mgh) 
k
2 

d must be positive, so d =
d=
1
mg + (mg)2 + 2kmgh
k
(
)
1
((1.20 kg)(9.80 m/s 2 ) +
1600 N/m
((1.20 kg)(9.80 m/s 2 ))2 + 2(1600 N/m)(1.20 kg)(9.80 m/s 2 )(0.80 m)
d = 0.0074 m + 0.1087 m = 0.12 m = 12 cm
EVALUATE:
It was important to recognize that the total displacement was h + d; gravity continues to
do work as the book moves against the spring. Also note that with the spring compressed 0.12 m it
exerts an upward force (192 N) greater than the weight of the book (11.8 N). The book will be
accelerated upward from this position.
7.25.
IDENTIFY:
Apply Eq.(7.13) and F = ma .
SET UP:
Wother = 0 . There is no change in U grav . K1 = 0 , U 2 = 0 .
EXECUTE:
1
2
kx 2 = 12 mvx2 . The relations for m, vx , k and x are kx 2 = mvx2 and kx = 5mg.
Dividing the first equation by the second gives x =
k = 25
vx2
, and substituting this into the second gives
5g
mg 2
.
vx2
(a) k = 25
(b) x =
(1160 kg)(9.80 m/s 2 )2
= 4.46 × 105 N/m
(2.50 m/s)2
(2.50 m/s)2
= 0.128 m
5(9.80 m/s 2 )
EVALUATE:
Our results for k and x do give the required values for ax and vx :
ax =
kx (4.46 × 105 N/m)(0.128 m)
k
=
= 49.2 m/s 2 = 5.0g and vx = x
= 2.5 m/s .
m
1160 kg
m
7.28.
IDENTIFY and SET UP:
The force is not constant so we must use Eq.(6.14) to calculate W. The
properties of work done by a conservative force are described in Section 7.3.
 
2 
W = ∫ F ⋅ dl , F = −α x 2 iˆ
1

EXECUTE:
(a) dl = dyĵ (x is constant; the displacement is in the + y-direction )


F ⋅ dl = 0 (since iˆ ⋅ ĵ = 0) and thus W = 0.

(b) dl = dxiˆ


F ⋅ dl = (−α x 2 iˆ) ⋅ (dxiˆ) = −α x 2 dx
W=
∫
x2
x1
(−α x 2 )dx = − 13 ax 3
x2
x1
= − 13 α (x23 − x13 ) = −
12 N/m 2
((0.300 m)3 − (0.10 m)3 ) = −0.10 J
3

(c) dl = dxiˆ as in part (b), but now x1 = 0.30 m and x2 = 0.10 m
W = − 13 α (x23 − x13 ) = +0.10 J
(d) EVALUATE:
The total work for the displacement along the x-axis from 0.10 m to 0.30 m and then
back to 0.10 m is the sum of the results of parts (b) and (c), which is zero. The total work is zero when
the starting and ending points are the same, so the force is conservative.
EXECUTE:
Wx →x = − 13 α (x23 − x13 ) = 13 α x13 − 13 α x23
1
2
The definition of the potential energy function is Wx →x = U1 − U 2 . Comparison of the two expressions
1
2
3
for W gives U = α x . This does correspond to U = 0 when x = 0.
1
3
EVALUATE:
In part (a) the work done is zero because the force and displacement are perpendicular. In
part (b) the force is directed opposite to the displacement and the work done is negative. In part (c) the
force and displacement are in the same direction and the work done is positive.
7.32.
IDENTIFY and SET UP:
Use Eq.(7.17) to calculate the force from U (x). Use coordinates where the
origin is at one atom. The other atom then has coordinate x.
EXECUTE:
6C
dU
d  C 
d  1
Fx = −
= −  − 66  = +C6  6  = − 76
dx
dx  x 
dx  x 
x
The minus sign mean that Fx is directed in the −x-direction, toward the origin. The force has
magnitude 6C6 / x 7 and is attractive.

EVALUATE:
U depends only on x so F is along the x-axis; it has no y or z components.
7.34.
IDENTIFY:
Apply F(x) = −
dU (x)
.
dx
d(1 / x)
1
=− 2
dx
x
d(−Gm1 m2 / x)
Gm1 m2
 d(1 / x) 
= Gm1 m2 
=−
EXECUTE:
Fx (x) = −
. The force on m2 is in the

dx
x2
 dx 
SET UP:
−x-direction . This is toward m1 , so the force is attractive.
EVALUATE:
By Newton's 3rd law the force on m1 due to m2 is Gm1 m2 / x 2 , in the +x-direction
(toward m2 ). The gravitational potential energy belongs to the system of the two masses.
7.38.
IDENTIFY:
Apply Eq.(7.16).
dU
SET UP:
is the slope of the U versus x graph.
dx
EXECUTE:
(a) Considering only forces in the x-direction, Fx = −
dU
and so the force is zero when
dx
the slope of the U vs x graph is zero, at points b and d.
(b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this
point is stable.
(c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic
energy, and the marble tends to move further away, and so d is an unstable point.
EVALUATE:
At point b, Fx is negative when the marble is displaced slightly to the right and Fx is
positive when the marble is displaced slightly to the left, the force is a restoring force, and the
equilibrium is stable. At point d, a small displacement in either direction produces a force directed away
from d and the equilibrium is unstable.
7.47.
(a) IDENTIFY:
Use work-energy relation to find the kinetic energy of the wood as it enters the rough
bottom.
SET UP:
Let point 1 be where the piece of wood is released and point 2 be just before it enters the
rough bottom. Let y = 0 be at point 2.
EXECUTE:
U1 = K 2 gives K 2 = mgy1 = 78.4 J.
IDENTIFY:
Now apply work-energy relation to the motion along the rough bottom.
SET UP:
Let point 1 be where it enters the rough bottom and point 2 be where it stops.
K1 + U1 + Wother = K 2 + U 2
EXECUTE:
Wother = W f = − µk mgs, K 2 = U1 = U 2 = 0; K1 = 78.4 J
78.4 J − µk mgs = 0; solving for s gives s = 20.0 m.
The wood stops after traveling 20.0 m along the rough bottom.
(b) Friction does −78.4 J of work.
EVALUATE:
The piece of wood stops before it makes one trip across the rough bottom. The final
mechanical energy is zero. The negative friction work takes away all the mechanical energy initially in
the system.
7.53.
IDENTIFY:
Use the work-energy theorem, Eq.(7.7). Solve for K 2 and then for v2 .
SET UP:
Let point 1 be at his initial position against the compressed spring and let point 2 be at the end
of the barrel, as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by
the 4400 N force.
K1 + U1 + Wother = K 2 + U 2
Take y = 0 at his initial position.
EXECUTE:
K1 = 0,
K 2 = 12 mv22
Wother = Wfric = − fs
Wother = −(40 N)(4.0 m) = −160 J
Figure 7.53
U1, grav = 0, U1,el = 12 kd 2 , where d is the distance the spring is initially compressed.
F = kd so d =
F
4400 N
=
= 4.00 m
k 1100 N/m
and U1,el = 12 (1100 N/m)(4.00 m)2 = 8800 J
U 2, grav = mgy2 = (60 kg)(9.80 m/s 2 )(2.5 m) = 1470 J, U 2,el = 0
Then K1 + U1 + Wother = K 2 + U 2 gives
8800 J − 160 J = 12 mv22 + 1470 J
2(7170 J)
= 15.5 m/s
60 kg
EVALUATE:
Some of the potential energy stored in the compressed spring is taken away by the work
done by friction. The rest goes partly into gravitational potential energy and partly into kinetic energy.
1
2
mv22 = 7170 J and v2 =
7.58.
IDENTIFY:
Conservation of energy says the decrease in potential energy equals the gain in kinetic
energy.
SET UP:
Since the two animals are equidistant from the axis, they each have the same speed v.
EXECUTE:
One mass rises while the other falls, so the net loss of potential energy is
(0.500 kg − 0.200 kg)(9.80 m/s 2 )(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the
animals and is equal to
1
2
mtot v 2 , and v =
2(1.176 J)
= 1.83 m/s.
(0.700 kg)
EVALUATE:
The mouse gains both gravitational potential energy and kinetic energy. The rat’s gain in
kinetic energy is less than its decrease of potential energy, and the energy difference is transferred to
the mouse.
7.74.
IDENTIFY:
Apply Eq.(7.14) to the motion of the package. Wother = W f , the work done by the kinetic
k
friction force.
SET UP:
f k = µk n = µk mg cosθ , with θ = 53.1∞. Let L = 4.00 m , the distance the package moves
before reaching the spring and let d be the maximum compression of the spring. Let point 1 be the
initial position of the package, point 2 be just as it contacts the spring, point 3 be at the maximum
compression of the spring, and point 4 be the final position of the package after it rebounds.
EXECUTE:
(a) K1 = 0 , U 2 = 0 , Wother = − f k L = − µk Lcosθ . U1 = mgLsin θ . K 2 = 12 mv 2 , where v is
the speed before the block hits the spring. Eq.(7.14) applied to points 1 and 2, with y2 = 0 , gives
U1 + Wother = K 2 . Solving for v,
v = 2gL(sin θ − µk cosθ ) = 2(9.80 m/s 2 )(4.00 m)(sin53.1° − (0.20)cos53.1°) = 7.30 m/s.
(b) Apply Eq.(7.14) to points 1 and 3. Let y3 = 0 . K1 = K 3 = 0 . U1 = mg(L + d)sin θ . U 2 = 12 kd 2 .
Wother = − f k (L + d) . Eq.(7.14) gives mg(L + d)sin θ − µk mg cosθ (L + d) = 12 kd 2 . This can be written
as d 2
k
− d − L = 0. The factor multiplying d 2 is 4.504 m −1 , and use of the
2mg(sin θ − µk cosθ )
quadratic formula gives d = 1.06 m .
(c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end
of the motion the spring has no potential energy, and the distance below the starting point is determined
solely by how much energy has been lost to friction. If the block ends up a distance y below the starting
point, then the block has moved a distance L + d down the incline and L + d − y up the incline. The
magnitude of the friction force is the same in both directions, µk mg cos , and so the work done by
friction is − µk (2L + 2d − y)mg cos . This must be equal to the change in gravitational potential
energy, which is −mgy sin . Equating these and solving for y gives
y = (L + d)
2 µk cosθ
2 µk
= (L + d)
. Using the value of d found in part (b) and the given
sin θ + µk cosθ
tan θ + µk
values for µk and θ gives y = 1.32 m .
EVALUATE:
Our expression for y gives the reasonable results that y = 0 when µk = 0 ; in the absence
of friction the package returns to its starting point.
7.77.
IDENTIFY:
ax = d 2 x/dt 2 , a y = d 2 y/dt 2 . Fx = max , Fy = ma y . U = ∫ Fx dx + ∫ Fy dy .
SET UP:
d
d
(cos ω 0 t) = −ω 0 sin ω 0 t .
(sin ω 0 t) = ω 0 cos ω 0 t .
dt
dt
1
∫ sin ω t dt = − ω
0
cos ω 0 t . vx = dx / dt , v y = dy / dt . E = K + U .
0
EXECUTE:
(a) ax = d 2 x/dt 2 = −ω 02 x, Fx = max = −mω 02 x.
a y = d 2 y/dt 2 = −ω 02 y, Fy = ma y = −mω 02 y
1
∫ cosω t dt = ω
0
0
sin ω 0 t ,
1
(b) U = −  ∫ Fx dx + ∫ Fy dy  = mω 02  ∫ xdx + ∫ ydy  = mω 02 (x 2 + y 2 )



 2
(c) vx = dx/dt = −x0ω 0 sin ω 0 t = −x0ω 0 ( y/y0 ). v y = dy/dt = + y0ω 0 cos ω 0 t = + y0ω 0 (x/x0 ).
(i) When x = x0 and y = 0, vx = 0 and v y = y0ω 0 ,
K=
1
1
1
1
m(vx2 + v 2y ) = my02ω 02 , U = ω 02 mx02 and E = K + U = mω 02 (x02 + y02 )
2
2
2
2
(ii) When x = 0 and y = y0 , vx = −x0ω 0 and v y = 0 ,
1
1
1
K = ω 02 mx02 , U = mω 02 y02 and E = K + U = mω 02 (x02 + y02 )
2
2
2
EVALUATE:
The total energy is the same at the two points in part (c); the total energy of the system is
constant.