Chapter 8 8.3 In Fig 8-30, a 2.00g ice flake is released from the edge ...

Chapter 8
8.3 In Fig 8-30, a 2.00g ice flake is released from the edge of a hemispherical bowl whose radius is
22 cm. The flake=bowl contact is frictionless. (a) how much work is done on the flake by the
gravitation force during the flake’s descent to the bottom of the bowl? (b) What is the change in
the potential energy of the flake Earth system during the descent? (c) If that potential energy is
taken to be zero at the bottom of the bowl, what is its value when the flake is released? (d) If
instead, the potential energy is taken to be zero at the release point, what is its value when the flake
reaches the bottom of the bowl (e) If the mass of the flake were doubled, would the magnitudes of
the answers to (a) through (d) increase, decrease, or remain the same?
a) The work done is
W = −ΔU
= −(U f
−U i
)
= −(0− mgy)
= 2 ×10 −3 ⋅ 9.8m / s 2 ⋅ 0.22m
= 4.312 ×10 −3 J
For this part of the problem, assume the bottom of the bowl is zero height
b) See part (a). ΔU = −4.312 ×10 −3 J
U f
= 0
U i
= mgR
W = −(U f
−U i
)
= mgR
c) See part (a) U i
= 4.312 ×10 −3 J
d) Assume that the top of the bowl is 0 height. The bottom would be -R
U i
= 0
U f
= mg(−R) = −4.312×10 −3 J
e) All of these answers are linear in m. If you double the mass, the results double.
8.5 In Fig. 8-32, a frictionless roller coaster car of mass m = 825kg tops the first hill with a speed
of v =17m / s at height h=42.0 m. How much work does the gravitational force do on the car
force that point to (a) point A, (b) point B, and (c) point C. If the gravitation potential energy of the
car-Earth system is taken to be zero at C, what is the value when when the car is at (d) B and (e) A?
(f) If the mass m were doubled, would the change in gravitational potential energy of the system
between points A and B increase, decrease, or remain the same?
We can work this problem using our knowledge of the relationship between work and the change in
potential energy.

W = −ΔU
U i
= mgh
a) At point A
b) At point B
W = −ΔU
U i
= mgh
U f
= mgh
W = −(U f
−U i
)
= −(mgh − mgh)
= 0
W = −ΔU
U i
= mgh
U f
= mg h 2
W = −(U f
−U i
)
= −(mg h 2 − mgh)
= mg h 42m
= 825kg ⋅ 9.8⋅
2 2
=169,785J
c) At point C
W = −ΔU
U i
= mgh
U f
= mg⋅ 0
W = −(U f
−U i
)
= −(0− mgh)
= 339,570J
d) The gravitational potential energy at B is
= mgh = 825⋅ 9.8 ⋅ 42m
e) The gravitational potential at A is
U B
= mg h 2
= 825 ⋅ 9.8⋅ 42m
2
=169,785J

U A
= mgh
= 825 ⋅ 9.8⋅ 42m
= 339,570J
f) If you double the mass, all results double.
8.7 In Fig. 8-33, a small block of mass m = 0.032kg can slide along the frictionless loop-the-loop
with a loop radius of R=12 cm. The block is released from rest at a point P, at height 5.0F above
the bottom of the loop. A block slides on a track from a height 5R.
a) How much work does the weight do on the block as it travels from P to Q? Assume that the
potential energy at the bottom is 0.
W = −ΔU
= −(mgy f
− mgy i
)
= −(mgR − mg⋅ 5R)
= 4mgR
= 4 ⋅ 0.032kg ⋅ 9.8m / s 2 ⋅ 0.12m
= 0.151J
b) How much work does the weight do on the block as it travels from P to the top of the loop?
Assume that the potential energy at the bottom is 0.
W = −ΔU
= −(mgy f
− mgy i
)
= −(mg ⋅ 2R − mg ⋅ 5R)
= 3mgR
= 3⋅ 0.032kg ⋅ 9.8m / s 2 ⋅ 0.12m
= 0.113J
c-e) Calculate the potential energies. Assume that the potential energy at the bottom is 0.
At point P U = mg(5R) = 0.188J
At point Q U = mg(R) = 0.0376J
At top of loop U = mg(2R) = 0.0753J
f) The potential energies and work done are unaffected by the initial velocity that the particle might
have.

8.18 A 2.0 kg block is dropped from a height of 40 cm onto a spring of spring constant of
k=1960N/m. Find the maximum distance that the spring is compressed.
The mass will stop when all of the change in gravitational potential energy has been converted into
energy in the spring. We can use energy conservation to do this problem.
Let’s consider the distance that the spring compresses. The energy that is stored in the spring is
where y is the amount of compression.
U s
= 1 2 k y 2
Defining zero height from the point where the spring is not stretched or compressed we can write
the initial energy potential as
U gi
= mgh
The energy when the block is dropped is equal to the energy at the bottom of the compression. The
block is dropped from rest so there is no initial kinetic energy. The energy at the bottom is the
compressed spring and a gravitational piece.
h
0
y

E i
= K i
+U i
= 0 + mgh
E f
= K f
+U f
= 0 + 1 2 k y2 + mgy
E f
=
E i
1
2 k y2 + mgy = mgh
1
2 k y2 + mgy − mgh = 0
−mg ±
y =
(−mg) 2 − 4 ⋅ 1 2 k ⋅(−mgh)
2 ⋅ 1 2 k = −0.1m
The negative sign is to be expected, since the final position is below the starting position.
8.23 The string in Fig 8-38 is L=120 cm long, has a ball attached to one end, and is fixed at its
other end. The distance d to the fixed peg at point P is 75.0 cm. When the initially stationary ball
is released wit the string horizontal as shown, it will swing along the dashed arc. What is its speed
when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg.
We begin by computing the total energy.
E i
= 1 2 mv 2 i
+ mgh i
= 0 + mgL
E i
= mgL
The total energy remains constant throughout this problem. We now consider the lowest point.
E Low
= 1 2 mv 2 Low
+ mgh Low
h Low
= 0
E Low
= 1 2 mv Low 2 + 0
E Low
= E i
1
2 mv 2 Low
= mgL
v Low
=
b) We now consider the high point after catching
2gL = 4.85m / s

E High
= 1 2 mv 2 High
+ mgh High
h High
= 2 ⋅ (L − d)
E High
= 1 2 mv High 2 + 2mg(L − d)
E High
= E i
1
2 mv 2 High
+ 2mg(L − d) = mgL
1
2 mv 2 High
= 2mgd − mgL
v High 2 = g(4d −2L)
v High
=
g(4d −2L)
= 2.42m / s
8.29 In fig 8.34, a 12 kg block is released from rest on a 30 degree frictionless incline. Below the
block is a spring that can be compressed 2.0 cm by a force of 270N. The block momentarily stops
when it compresses the spring by 5.5 cm. (a) How far does the block move down the incline from
its rest position to this stopping point. (b) What is the speed of the block just as it touches the
spring.
(a) All of the potential energy of the spring comes from change in gravitational potential energy.
k =
270N
0.02m =1.35 ×10 4 N / m
U s
= 1 2 k x 2
= 1 2 ⋅1.35 ×104 N / m⋅(0.055) 2
= 20.42J
U s
= ΔU g
= mgΔy
Δy =
U s
mg = 20.42J
12kg ⋅ 9.8 = 0.1736m
d =
Δy
sinθ = 0.1736m
sin 30 = 0.3472m
(b) To compute the kinetic energy at impact, we need to find the vertical distance that the block
moved. The distance down the incline before compressing the spring can be computed.
′ d = d − 0.055m
= 0.3472m − 0.055m
= 0.2922m

Since we know the distance down the incline, we can find the vertical drop, change in potential
energy and finally the kinetic energy and velocity.
Δ y ′ = − d ′ sinθ
= −0.2922⋅ sin30
= −0.1461m
K f
− K i
=W = −ΔU
1
2 mv 2 f
− 1 2 mv 2 i
= −mgΔy
1
2 mv 2 f
−0 = −mgΔy
v f
=
−2gΔy
= −2⋅ 9.8m / s 2 ⋅ −0.1461m
=1.692m / s
8.36 A boy is seated on top of a hemispherical mound of ice of radius R=13.8m. He begins to
slide down the ice, with a negligible initial speed. Approximate the ice as being frictionless. At
what height does the boy lose contact with the ice. Show that he leaves the ice at point whose height
is 2R/3 if the ice is frictionless.
We begin with a picture.
We are interested in finding when the Normal force disappears. We treat this as a circular motion
problem
mv 2
R
= mgcosθ − N
N = 0
mv 2
R
= mgcosθ
v =
Rgcos θ

We now know the velocity when the boy comes off the ice. We now use conservation of energy to
find out how high he is when this happens.
E i
= 1 2 mv i 2 + mgh i
= 0 + mgR
E f
=
1 2 mv 2 f
+ mgh f
E f
=
E i
1
2 mv 2 f
+ mgh f
= mgR
v f 2 = Rg cosθ
cosθ =
h f
R
1
2 mR g ⋅ h f
R + m gh f
= mgR
h f
= 2 3 R = 9.2m
8.71 A conservative force F(x) acts on a 2.0 kg particle that moves along an x axis. The potential
energy U(x) associated with F(x) is graphed below. When the particle is at x=2.0 m, its velocity is
-1.5 m/s. What are the (a) magnitude and (b) direction of F(x) at this position? Between what
positions on the (c) left and (d) right does the particle move? (e) What is the particle’s speed at
x=7.0 m.

a and b) The magnitude and direction of the force at x=2 are given by sign and magnitude of the
slope at x=2. F=-slope of U(x) vs. x.
At x=2, the slope = -4.667, so F = 4.667 (positive indicates that the force is in the +x direction)
c and d) To find the turning points, we need to find where the total energy line crosses U(x). At
these crossing points, all of the energy is potential. This means that the particle must stop (since it
has zero kinetic energy) and turn around.
At x=2
KE = 1 2 mv2 = 1 2 ⋅ 2.0kg ⋅(−1.5m / s)2 = 2.25J
U(2) = −7.667( fromgraph)
E = KE +U = −5.417J
The graph below shows the total energy line (dashed). The turning points are where the total
energy line crosses the U(x) There are at approximately 1.5 m and 13.75 m

e) We can use the total energy to find the particle’s speed at x=7m
KE =
1 2 mv2
U(2) = −17J ( fromgraph)
E = KE +U = −5.417J
KE = E −U
1
2 mv2 = E −U
2(E −U) 2(−5.14 − (−17J ))
v =
= = 3.4m / s
m
2.0kg
You may get a slightly different answer because of the inherent problems in reading off the graph.