Ch 8, Potential Energy and Conservation of Energy Homework: Ch8; 9, 11, 13, 21, 26, 31, 52, 57
Questions 3, 7, 8, 9, 10, 13
Additional problems from Serway's fourth edition
(4 ed) 8.2 A 3.0-kg block starts at a height h = 60 cm = 0.60 m on a
plane that has an inclination angle of 30o as in Figure P8.20.
Upon reaching the bottom, the block slides along a horizontal surface. If
the coefficient of friction on both surfaces is 
= 0.20, how far does the block slide on the horizontal surface before coming
to rest?
(Hint: Divide the path into two straight-line parts.)
(a) the change in its kinetic energy,
(b) the change in its potential energy, and
(c) it speed at x = 5.0 m if it starts from rest at x = 2.0 m
Conceptual Questions
Q8.3 A bowling ball is suspended from the ceiling of lecture hall by a strong cord. The bowling ball is drawn away from its equilibrium position and released from rest at the top of a student's nose as shown in Figure Q8.3. If the student remains stationary, explain why she will not be struck by the ball on its return swing. Would the student be safe if she pushed the ball as she released?
As the bowling ball swings back and forth energy is conserved. Energy changes from gravitational potential energy to kinetic energy and back again and back and forth. But the total energy remains constant. When the student releases the bowling ball from rest it has no kinetic energy so its total energy is equal to the gravitational potential energy at that point. It swings out, stops (meaning its KE is again zero) and then starts back. At the bottom of its swing it has minimum gravitational potential energy and maximum kinetic energy. As it swings on -- toward the student -- it is slowing down or loosing kinetic energy as it gains height and gravitational potential energy. It will completely stop when its gravitational potential energy equals its total energy -- at that is exactly at the student's chin.
However, if the student pushes the ball that means she gives it some initial kinetic energy in addition to its initial gravitational potential energy. It will always continue on until its gravitational potential energy just equals its total energy. This time it will stop at a greater height since it has a larger amount of total energy. Stopping at a greater height means it will run into the student's chin!
Q8.7 If three conservative forces and one nonconservative force act on a system, how many potential energy terms appear in the equation that describes this system?
We can write a potential energy term for each conservative force so there will be three potential energy terms. We can not write a potential energy term for a nonconservative force. Fricition is a ready example of such a nonconservative force.
Q8.8 Consider a ball fixed to one end of a rigid rod whose other end pivots on a horizontal axis so that the rod can rotate in a vertical plane. What are the positions of stable and unstable equilibrium?
Stable equilibrium will be with the rod at the bottom of its arc (or possible arc).
Unstable equilibrium will be wit the rod at the top of its arc.
Q8.9 Is it physicall possible to have a situation where E - U < 0?
The total energy E is the sum of the potential energy U and the kinetic energy K,
E = U + K
Therefore, the kinetic energy is the total energy minus the potential energy,
K = E - U
Since K = (1/2) m v2, the kinetic energy is always positive. Therefore, it is not possible to have a situation where
E - U < 0
Q8.10 What would the curve of U versus x look like if a particle were in a region of neutral equilibrium?
Neutral equilibrium means zero net force. The force is the slope of the U vs x curve. That means the slope is zero so the curve is a straight, horizontal line.
Q8.13 If only one external force acts on a particle, does it necessarily change the particle's
(a) kinetic energy?
Yes.
(b) velocity?
Yes (just think of Newton's Second Law, F = m a).
8.9 A single conservative force acting on a particle varies as F = ( - A x + B x2) i N, where A and B are constants and x is in meters.
(a) Calculate the potential energy associated with this force, taking U = 0 at x = 0.
(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 2.0 m to x = 3.0 m.
U(2) = (1/2) A (2)2 - (1/3) B (2)3 = 2 A - (8/3) B U(3) = (1/2) A (3)2 - (1/3) B (3)3 = (9/2) A - 9 B
U = Uf - Ui = U(3) - U(2) = [ (9/2) A - 9 B ] - [ 2 A - (8/3) B ]
E = K + U = constant
8.11 A 3.0-kg mass starts from rest and slides a distance d down a frictionless 30o incline, where it contacts an unstressed spring of negligible mass as in Figure P8.30. The mass slides an additional 0.20 m as it is brought momentarily to rest by compressing the spring (k = 400 N/m). Find the initial separation d between mass and spring.
In describing the initial potential energy, what do we want to choose as the reference position? While there are other good choices, I chose to make the potential energy zero when the mass stops after compressing the spring. That is, I choose the reference point for the potential energy -- the location at which the potential energy is zero -- to bed + 20 cm along the plane.
That means the block starts at an initial height h above this reference point. And that height is the "opposite side" of a right triangle so that
h = (d + 20 cm) sin 30o h = (d + 20 cm) (0.500)
h = 0.5 d + 10 cm
h = 0.5 d + 0.10 m
Initially,
Ei = Ki + Usi + Ugi Ei = 0 + 0 + m g h
Ei = (3.0 kg) (9.8 m/s2) (0.5 d + 0.10 m)
Ei = 14.7 d (kg-m/s2) + 2.94 J
Ef = Kf + Usf + Ugf
Ef = 0 + (1/2) k x2 + 0
Ef = (1/2) (400 N/m) (0.20 m)2
Ef = 8 J
Ei = Ef
Ei = 14.7 d (kg-m/s2) + 2.94 J = 8 J = Ef
14.7 d (kg-m/s2) = 5.1 J
d = 0.34 m
d = 34 cm
8.13 A particle of mass m = 5 kg is released from point A and slides on the frictionless track shown in Figure P8.13. Determine(a) the particle's speed at poins B and C and
(b) the net work done by the force of gravity in moving the particle from point A to point C.
[[ Fig P8.13 ]]
ETot = KE + U
KE = (1/2) m v2
U = m g h
ETot, A = 0 + (5 kg)(9.8 m/s2)(5 m) = 245 J
ETot = 245 J = constant
ETot, B = (1/2)(5 kg)(vB2) + (5 kg)(9.8 m/s2)(3.2 m) = ETot = 245 J
(1/2)(5 kg)(vB2) + 156.8 J = 245 J
(1/2)(5 kg)(vB2) = 245 J - 156.8 J = 88.2 J
vB2 = (35.38)(J/kg)
What about the units? Let's check and see.
vB2 = (35.38)(m2/s2)
vB = 5.9 m/s
Of course, finding the speed at point C is exactly the same,
ETot, C = (1/2)(5 kg)(vC2) + (5 kg)(9.8 m/s2)(2.0 m) = ETot = 245 J
(1/2)(5 kg)(vC2) + 98 J = 245 J
(1/2)(5 kg)(vC2) = 245 J - 98 J =147 J
vC2 = 147 J / 2.5 kg
vC2 = (147/2.5)(J/kg)
vC2 = (58.8)(J/kg)
vC = 7.7 m/s
WAC = m g
8.21 Two masses are connected by a light string passing over a light frictionless pulley as shown in Figure P8.17. The 5.0-kg mass is released from rest.
Using conservation of energy,
(a) determine the speed of the 3.0-kg mass just as the 5.0-kg mass hits the ground.
This is essentially the same as 8.10:Ki = 0 Ui = m1 g h1 + m2 g h2
Ui = (5.0 kg) (9.8 m/s2) (4.0m) + (3.0 kg) (9.8 m/s2) ( 0 )
Ui = 196 J
Ei = Ui + Ki = 196 J
E = constant
Ef = Ei
Ef = Kf + Uf
Kf = Ef - Uf
Uf = (5.0 kg) (9.8 m/s2) ( 0 ) + (3.0 kg) (9.8 m/s2) (4.0 m)
Uf = 117.6 J
Kf = 196 J - 117.6 J
Kf = 78.4 J
Kf = (1/2) (5.0 kg) v2 + (1/2) (3.0 kg) v2
Kf = (1/2) (8.0 kg) v2
(1/2) (8.0 kg) v2 = 78.4 J
v2 = 19.6 m2 / s2
v = 4.43 m/s
Note that both masses have the same speed.
(b) Find the maximum height to which the 3.0-kg rises.
At a height of 4.0 m, the 3.0-kg mass is moving upwards with a speed of 4.43 m/s. The 5.0-kg mass has just hit the ground so the cord goes loose and there is no other force on the 3.0-kg mass but the force of gravity. How much higher does it go? We know its acceleration so this is a kinematics problem from long ago. Perhaps the most efficient way is just to usev2 = vi2 + 2 a (s - si) a = - g = - 9.8 m/s2
si = 4.0 m
v = 0
vi = 4.43 m/s
v2 = vi2 + 2 a (s - si)
0 = (4.43 m/s)2 + 2 ( - 9.8 m/s2) ( s - 4.0 m)
0 = 19.6 m2/s2 - (19.6 m/s2) ( s - 4.0 m)
(19.6 m/s2 ) ( s - 4.0 m) = 19.6 m2/s2
s - 4.0 m = 1.0 m
s = 5.0 m
Of course, we could also solve this part of the problem using Conservation of Energy. At a height of 4.0 m, the 3.0-kg mass is moving upwards with a speed of 4.43 m/s. The 5.0-kg mass has just hit the ground so the cord goes loose and there is no other force on the 3.0-kg mass but the force of gravity. How much higher does it go?
For this part of the problem, the "initial conditions" of the 3.0-kg mass are just
vi = 4.43 m/s Ki = (1/2) m vi2 = (0.5)(3.0 kg)(4.43 m/s)2 = 29.4 J
Ui = m g yi = (3.0 kg)(9.8 m/s2)(4.0 m) = 117.6 J
Ei = Ki + Ui = 147 J
Ef = Kf + Uf
Kf = 0
because the block stops at its highest point.
Uf = Ef = Ei = 147 J Uf = m g yf = 147 J
(3 kg)(9.8 m/s2) yf = 147 J
yf = 5.0 m
And, of course, that is exactly what we found from the "kinetmatics approach". The two methods are essentially identical.
8.26 After its release at the top of the first rise, a roller-coaster car moves freely with negligible friction. The roller coaster showin in Fig P8.26 has a circular loop of radius 20.0 m. The car barely makes it around the loop: At the top of the loop, the riders are upside down and feel weightless.(a) Find the speed of the roller coaster car at the top of the loop (position 3).
Find the speed of the roller coaster car
(b) at position 1 and
(c) at position 2.
(d) Find the difference in height between positions 1 and 4 if the speed at position 4 is 10.0 m/s.
[[ Fig P8.26]]
Finding the speed at the top of the loop is a centripetal force problem, straight out of Chapter 6 on Circular Motion. Look back at homework problem 6.19.
With the pail of water in problem 6.19, just as with the roller coaster car here, the force of gravity supplies the centripetal force.
Fc = Fg = w = mg
Fc = m v2/r = m g
v2= g r
v2 = (9.8 m/s2) (20 m)
v2 = 196 m2/s2
v = 14 m/s
v3 = 14 m/s
This is v3 the speed at position 3 -- the top of the loop.
While not asked for in the text, let's find out the minimum height -- call it position 0 -- from which the roller coaster must start so it can get to this position 3 -- the top of the loop. Now, this -- and the rest of the problem -- just makes use of Energy Conservation.
E = KE + 0
E = (1/2) m v2 + m g h
E0 = 0 + m g h0
E3 = (1/2) m v32 + m g h3
E0 = m g h0 = (1/2) m v32 + m g h3 = E3
h0 = (1/2) v32 /g + h3
h0 = (1/2)(14 m/s)2/(9.8 m/s2) + 40 m
Remember, h3 is the height at the top of the loop so that is h3 = 2 r = 2 (20 m) = 40 m.
h0 = 50 m
Now, back to the details asked for in the textbook. What is the speed at position 1, the bottom of the loop? Again, we will use Conservation of Energy.
E = constant
E1 = E3
E1 = (1/2) m v12 + m g h1
E3 = (1/2) m v32 + m g h3
(1/2) m v12 + m g h1 = (1/2) m v32 + m g h3
(1/2) v12 + g h1 = (1/2) v32 + g h3
(1/2) v12 = (1/2) v32 + g h3 - g h1
(1/2) v12 = (1/2) v32 + g (h3 - h1)
v12 = v32 + 2 g (h3 - h1)
v12 = (14 m/s)2 + 2 (9.8 m/s2)(40 m)
v12 = 980 m2/s2
v1 = 31.3 m/s
Finding the speed at position 2 is exactly the same;
E = constant
E2 = E3
E2 = (1/2) m v22 + m g h1
E2 = (1/2) m v32 + m g h3
(1/2) m v22 + m g h2 = (1/2) m v32 + m g h3
(1/2) v22 + g h2 = (1/2) v32 + g h3
(1/2) v22 = (1/2) v32 + g h3 - g h2
(1/2) v22 = (1/2) v32 + g (h3 - h2)
v22 = v32 + 2 g (h3 - h2)
v22 = (14 m/s)2 + 2 (9.8 m/s2)(20 m)
v22 = 588 m2/s2
v2 = 24.3 m/s
8.31 The coefficient of friction between the 3.0-kg mass and surface in Figure P8.31 is 0.40. The system starts from rest. What is the speed of the 5.0-kg mass when it has fallen 1.5 m?
Ei = Ki + Ui Ei = 0 + m2 g h2
Ei = 0 + (5.0 kg) (9.8 m/s2) (1.5 m)
Ei = 73.5 J
This amount of energy goes into the kinetic energy of the system -- the kinetic energy of both masses -- and the heat associated with the work due to friction.
Kf = (1/2) m1 v2 + (1/2) m2 v2 = (1/2) (m1 + m2) v2 Kf = (1/2) (8 kg) v2 = (4 kg) v2
Uf = 0
Ef = Ei + Wf
Wf = - Ff s
Ff =
Fn = m1 g
Fn = (3.0 kg) (9.8 m/s2)
Ff = (0.40) (29.4 N) = 11.8 N
Wf = - (11.8 N) (1.5 m)
Wf = - 17.6 J
Ef = 73.5 J - 17.8 J = 55.7 J
Ef = Kf = (4 kg) v2 = 55.7 J
v2= 13.9 m2/s2
v = 3.7 m/s
8.46 A hollow pipe has one or two weights attache to its inner surface as shown in Figure P8.46. Characterize each configuration as being stable, unstable, or neutral equilibrium and explain each of your choices.
(a) is stable because a small movement will raise the center of mass and increase the potential energy of the system.(b) is in neutral equilibrium because the height of the center of mass will be unchanged by a small movement and, so, the potential energy of the system will be unchanged.
(c) is unstable because a small change will lower the center of mass and, thus, decrease the potential energy of the system.
8.52 A 200-g particle is released from rest at point A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Figure P8.52). Calculate(a) its gravitational potential energy at point A relative to point B,
(b) its kinetic energy at point B,
(c) its speed at point B, and
(d) its kinetic energy and potential energy at point C.
(a) At point A, the height ish A = R so the potential energy is
UA = m g hA = m g R (b) At point B, the height is zero,
hB = 0 so the potential energy is zero. This means the total energy is all kinetic energy,
UB = 0 EB = KB + UB = m g R = EA
KB = m g R
(c) We know kinetic energy is given by
KB = (1/2) m vB2 = m g R v B2 = 2 g R
v B = SQRT(2 g R)
(d) At point C, the vertical distance is
hC = (2/3) R That means
UC = m g hC = (2/3) m g R Since the bowl is frictionless, we know energy is conserved. This means the energy at point C is the same as the initial energy at point A;
EC = KC + UC = EA
KC = EA - UC
KC = m g R - (2/3) m g R
KC = (1/3) m g R
Solutions to the additional problems from Serway's fourth edition.
(4 ed) 8.1 A 5.0-kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 3.5-kg mass as shown in Figure P8.*. Use conservation of energy to determine the final speed of the 5.0-kg mass after it has fallen (starting from rest) 2.5 m.
Ki = 0
Ui = m1 g h1 + m2 g h2
Ui = (5.0 kg) (9.8 m/s2) (2.5 m) + (3.5 kg) (9.8 m/s2) ( 0 )
Ui = 122.5 J
Ei = Ui+ Ki = 122.5 J
E = constant
Ef = Ei
Ef = Kf + Uf
Kf = Ef - Uf
Uf = (5.0 kg) (9.8 m/s2) ( 0 ) + (3.5 kg) (9.8 m/s2) (2.5 m)
Uf = 87.25 J
Kf = 122.5 J - 87.25 J
Kf = 35.25 J
Kf = (1/2) (5.0 kg) v2 + (1/2) (3.5 kg) v2
Kf = (1/2) (8.5 kg) v2
(1/2) (8.5 kg) v2 = 35.25 J
v2 = 8.3 m2 / s2
v = 2.9 m/s
Note that both masses have the same speed.
= 0.20, how far does the block slide on the horizontal surface before coming to
rest?(Hint: Divide the path into two straight-line parts.)
Ei = Ui + Ki Ei= m g hi + 0
Ei = (3 kg) (9.8 m/s2) (0.60 m)
Ei = 17.64 J
On the way down, that itinital potential energy is changed into kinetic energy and work done against friction Wf . The kinetic energy at the bottom of the incline Kbot is found byEi + Wf = Ebot Ei = Ebot - Wf = Kbot + Ubot - Wf
Kbot = Ei - Ubot + Wf
Be (very!) careful of the sign on Wf and where you place it; is this the work done by friction on the block or is the work done on the plane by the block? You can use either -- if it is clear to you which you are using and which sign that interpretation requires. To be consistent with Serway's text, we will take Wf to be the work done by friction on the block which means Wf is intrinsically negative.
W f = - F f s From the diagram, we find
sin 30o = 0.500 = opp/hyp s = hyp = opp/0.5 = 0.60 m/0.5
s = 1.2 m
Ff =
Also, from the diagram, we find that
Fn = m g cos 30o = (3 kg) (9.8 m/s2) (0.866) Fn = 25.5 N
Ff = (0.20) (25.5 N) = 5.1 N
W f = - F f s
W f = - (5.1 N) (1.2 m)
W f = - 6.11 J
Kbot = Ei - Ubot + Wf
Kbot = 17.64 J - 0 - 6.11 J
Kbot = 11.53 J
This is the kinetic energy at the bottom. But, since that is our reference position for the potential energy, the potential energy at the bottom is zero. So this is the total energy -- the total, large-scale, mechanical energy -- of the system. 6.11 J of initial energy has gone into heat as friction did work on the block and plane.
Now, on the horizontal surface, the normal force Fn and the friction force Ff have changed;Fn = m g = (3.0 kg) (9.8 m/s2) Fn = 29.4 N
Ff = Fn = (0.20) (29.4 N) = 5.9 N
Wnet = Wf = - Ff shor
Wnet =
- Ff shor = - Kbot
(5.9 N) shor = 11.5 J
shor = 1.96 m
(4 ed) 8.3 A toy consists of a piece of plastic attached to a spring as in Figure P8.*. The spring is compressed 2.0 cm and the toy is released. If the mass of the toy is 100 g and it rises to a maximum height of 60 cm, estimate the force constant of the spring.
Ei = Usi = (1/2) k x2 = (1/2) k (0.02 m)2 Later, at the top of its "jump", when its velocity is momentarily zero, there is only gravitational potential energy,
Ef = Ugf = m g hf = (0.1 kg) (9.8 m/s2) (0.60 m) = 0.59 J Since we expect energy to be conserved, this means
Ei = Ef (1/2) k (0.02 m)2 = 0.59 J
k = 2 940 N / m
(4 ed) 8.4 A particle of mass m starts from rest and slides down a frictionless track as in Figure P8.*. It leaves the track horizontally, striking the ground as indicated in the sketch. Determine h.
What is the horizontal velocity of the ball as it leaves the track? How long does it take (how much time is required) for the ball to fall 1.25 m?y = yo + vyo t + (1/2) ay t2 - 1.25 m = 0 + 0 + (1/2) ( - 9.8 m/s2) t2
t2 = 0.255 s 2
t = 0.505 s
During this time, the ball travels 1.0 m so its horizontal speed must be
v = vx = 1.0 m /0.505 s v = 1.98 m/s
That means its kinetic energy at the bottom of the track is
K = (1/2) m v2 = (1/2) m (1.98 m/s)2 At the bottom of the track, the gravitational potential energy is zero; that is, we have choosen the bottom of the track as our reference point. That means the total energy is just the kinetic energy,
Ebot = K bot + U bot = K bot + 0 = K bot = (1/2) m (1.98 m/s)2 At the top of the track, when the ball was at rest, the kinetic energy was zero
Etop = Ktop + Utop = 0 + Utop = m g h Energy conservation means
Etop = Ebot m g h = (1/2) m (1.98 m/s)2
h = 1.96 m
(4 ed) 8.4 A 4.0-kg particle moves along the x-axis under the influence of a single conservative force. If the work done on the particle is 80.0 J as it moves from the point x = 2.0 m to x = 5.0 m, find(a) the change in its kinetic energy,
(b) the change in its potential energy, and
(c) it speed at x = 5.0 m if it starts from rest at x = 2.0 m
W = Wnet = 80 J W = Wnet =
K(x=2) = 0
K(x=5) = Kf = 80 J
K(x=5) = (1/2) m v2 = 80 J
(1/2) (4 kg) v2 = 80 J
v2 = 40 m2 / s2
v = 6.32 m / s
(c) Doug Davis, 2001; all rights reserved
