Chapter 7: WORK & ENERGY II Hints: Updated 4/25/12
1. The work-energy principle can be expressed mathematically in different ways. Convince yourself
that the following pairs of equations are equivalent expressions.
a)
b)
c)
d)
[Wnet=K ] and [Wnon-cons=K + U]
[U + K = Uo + Ko + Wnon-cons] and [Wnon-cons=K + U]
[K = - U] and [U + K = Uo + Ko]
How do you decide whether to apply the “work-energy principle” or “the energy-conservation
principle” to a problem?
a) Wnet = Wnon-cons + Wcons = Wnon-cons +(-∆U), etc…
b)& c) ∆K=K-Ko and ∆U=U-Uo, etc…
e) You look at all the forces in the “system” and determine whether there is some net work done by nonconservative forces, if the answer is No, then the total mechanical energy (TME) is conserved.
2. A 2 kg mass slides from rest down a frictionless inclined plane of angle . It slides 0.5 m until it
makes contact with a spring of constant 120 N/m.
S
a) Determine the velocity of the mass when it
0.5 m
makes contact with the spring. Explain why this
is not the maximum velocity of the mass.
b) Determine the maximum velocity of the mass. Hint: First determine the point where the acceleration
is zero.
c) Determine the maximum compression of the spring. Determine the acceleration at that point.
d) Now assume that there was kinetic friction all along with a coefficient of 0.125. Find the maximum
compression of the spring.
e) Determine the maximum distance that the mass slides up the ramp after bouncing back from the
maximum compression in (d).
a) Since the incline is frictionless, Utop=Kbottom v =2.45 m/s. This speed is not the maximum because
the block continues to accelerate down the incline while it compresses the spring until the force from
the spring becomes larger that the component of the weight down the incline. At that point the mass
will start to slow down, still compressing the spring, until it comes to rest, before bouncing back.
b) The acceleration is zero when Fspring=ksequil=mgsinø. This happens partly through the compression
of the spring when sequil =0.10 m. Then using energy conservation you can show that vmax=2.57 m/s.
c) At this point the mass is instantaneously at rest and all the initial Ug (relative to that point) has
changed to Us: mg(0.5 +smax)sinø =ksmax2/2  Solving this quadratic give smax=0.43 m. Acceleration
at that point (determined from Fnet=ma) is 20 m/s2, directed up the incline. This acceleration is what
makes the block bounce back.
d) Rework the problem using the work-energy principle (Wnon-cons=K +U)I got smax=0.38 m.
e) Use the work-energy principle. I got 0.63 m altogether or 0.35 m beyond the relaxed spring position.
3. Consider a roller coaster with a circular loop-the-loop where friction is negligible. The mass of the
cart and passengers is M. Using the given information, determine the following:
g
ø
H
R
h
a) The minimum initial height H so that a cart at rest at the top of the hill will be able to round the top
of the loop.
b) Determine the speed of the cart at the bottom of the loop and half way up to the top.
c) As a review of circle dynamics, also determine the contact, normal force at the bottom, the top, and
the middle of the loop.
d) Now assume that the height of the hill is only 2R (not enough to complete the loop), show that the
cart leaves the track when it is at an angle of 48o with the vertical.
e) How high did the cart rise in (d) in terms of R?
a) This was done in class, check your notes. Compare the potential energy at the top of the hill to the
energy at the top of the loop, where vmin=√(Rg) Hmin=2.5R.
b) Using energy conservation vbottom=(5Rg)1/2; vmiddle=(3Rg)1/2
c) Review the dynamics of vertical loop-the-loop: Fc=N+mgcosø Nbottom=6mg; Nmiddle=3mg; Ntop=0
d) Recall that in the loop-the-loop, the speed when the cart loses contact (N=0) is v=(Rgcosø)1/2. The
height at that point is determinde by the geometry to be h=(R+Rcosø) from the ground. Use energy
conservation to determine that the angle value ø=48º.
e) R(1+cos48ºR
4. A skier starts from rest at the top of a circular hill of radius R. At some point the skier loses contact
with the curve and becomes a projectile until it lands some distance from the bottom of the hill.
Loses
a) Show that he loses contact with the hill when he has
contact
h
moved downward a vertical distance of R/3.
ø=48º
R
b) In what direction with respect to the horizontal is he moving
ø
when he becomes a projectile?
c) As a review of projectile motion, determine how far the skier
Landing
d
lands from the bottom of the hill. Assume the radius is 3 m.
spot
a) Skier drops a vertical distance h =(R-Rcosø). We also know that when the normal force is 0, the
speed in the circle is v=√Rgcosø). Putting this information into the energy conservation equation
you can show that the angle is º and that the vertical drop h=R/3.
b) By geometry the direction is 48ºbelow the horizontal.
c) Determine the x and y components of the velocity of the skier. The vertical displacement is 2R/3. Use
the projectile kinematics formula for vertical displacement to determine the time in the air. Then use
the time to determine the horizontal displacement. I got tair= 0.38 s, skier travels horizontally 1.15 m,
and d=0.39 m.
5. To show the advantage of the energy method under the appropriate conditions,
consider a pendulum of length L swinging from an angle ø to the bottom.
ø
a) Explain why the tension doesn’t do any work on the bob, and
why energy is conserved as long as air resistance is negligible.
b) Use the energy conservation principle to determine the speed of
(L-Lcosø)
the bob at the bottom.
c) Determine the tension when the bob is initially released and when the bob is at the bottom.
d) Determine the tangential acceleration along the arc and explain why it isn’t constant.
e) It is also possible to determine the speed at the bottom by integrating the acceleration. To do
this you have to relate an infinitesimal piece of the arc (ds) to the angle that it subtends (d).
Show that you get the same answer as in (b). This demonstrates the advantage of the energy
method over kinematics for cases where the acceleration is not constant.
a) Tension is perpendicular to the path at every instant so the only force doing work is “mg”…
b) Geometry shows that the bob drops a vertical height of (L-Lcosø); so, vbottom=[2gL(1-cosø)] 1/2.
c) In general Fc=(T-mgcosø) and Ft=(mgsinø). At point of release Fc=0, so, To=mgcosø. At the
bottom: Tbot=mg +Fc =mg(3-2cosø).
d) Recall that atangential=gsinø. Clearly it isn’t constant because the angle changes from ø to zero.
e) Start with definition of acceleration and the chain rule to set up an integral for an infinitesimal part
of the swing: at=dv/dt=(dv/ds)(ds/dt)=(dv/ds)v ∫atds =∫gsinø(Ldø) = ∫vdv. Solving this integral
should give you the same answer as in (b).
6. A mass on a string is swung in a complete vertical circle of radius R.
a) Show that the difference between the magnitude of the
centripetal force at the bottom and at the top equals four times the
weight (4mg).
b) If the speed at the top is the minimum possible, what’s the speed
at the bottom?
c) Show that the difference between the magnitude of the
tension at the bottom and at the top equals six times the weight (6mg).
d) A stunt pilot flies his plane into a vertical circle. If he “feels weightless” at the top, how heavy
does he feel at the bottom of the circle?
a) Use energy conservation. Note that kinetic energy and centripetal force are easily related (Fc=2K/R).
b) The solution is the same as the earlier “roller-coaster” problem: vbottom= (5Rg)1/2
c) Relate the kinetic energies to the centripetal force at the top and the bottom: Tbottom=mg +Fc , and
Ttop= Fc - mg  Tbot - Ttop= (mvbot2/R) +(mvtop2/R)= 6mg.
d)This problem is basically the same as (c) but in a different context. The pilot is pushed toward the
center of the circle by the plane, so the normal force on the pilot from the plane is equivalent to the
tension in the string above and he would feel 6 times his weight at the bottom.
7. A 200 g block is pushed against a horizontal spring of constant 200 N/m until it is compressed 15
cm. When the mass is loses contact with the spring it moves over a horizontal rough patch of surface of
length 50 cm and coefficient of friction 0.2. After passing this patch the block continues to slide up a
frictionless ramp of angle 30o.
a) How far does the block rise on the ramp after the first pass?
b) The mass comes down and goes over the rough patch again
0.15 m
0.50 m
on the way back to the spring. How many passes will the
mass make over the patch before running out of energy?

c) Where along the rough patch will the block finally stop?
d) How long would the patch have to be so that the block never reaches the ramp?
a) From the work-energy principle: mgh=ks2/2 - mgx. I got h=1.03 m, which is the vertical height
after one pass.
b) Divide the initial U of the spring (2.25J) by the heat dissipated during one pass (0.2 J) over the
rough section. The block makes 11.25 passes.
c) After 11 passes the mass stops after ¼ of a pass, so it comes to rest 50/4=37.5 cm from the right end
of the rough patch (or 12.5 cm from the left end).
d) It would require (11.25x50)=562.5 cm=5.625 m of rough surface to stop the mass.
8. When two bodies are connected with a string over a pulley of negligible mass and friction, it is
possible to use the energy conservation principle (even though there is the non-conservative force of
tension) as long as it is applied to both bodies as a closed system. Consider the following two
examples:
Example 1 involves two unequal masses (M>m) hanging over a pulley with M
starting a distance h higher than m.
a) Explain why the work done by the tension (a non-conservative force)
doesn’t invalidate the energy-conservation principle here.
M
h
m
b) Determine the speed of each mass in terms of the masses and the distance
they move (without finding the acceleration) assuming they start from rest.
c) What does the work done by the tension on each mass represent in this problem?
Example 2 involves a mass M connected to a spring (constant k) over a pulley.
d) If you release the mass from rest with the spring unstretched,
how far will it fall before it is again (instantaneously) at rest?
e) How far has the mass fallen when it has its maximum velocity?
f) If the system were at equilibrium how much lower would the mass be?
M
h
k
a) The tension does “equal and opposite work” on each mass so the net work by non-cons. forces is 0.
b) This was done in class, check your notes. Energy is conserved for the system of the two masses. M
loses Ug and m gains Ug, while both masses gain K vfinal =[2(M-m)gh/(M+m)] 1/2.
c) WT =(+Th) on one mass and (-Th) on the other. These quantities represent the energy transferred
from the large mass to the smaller mass during the motion.
d) Ug =Us h= 2Mg/k
e) K is max at the half-way point to the max stretch: s’=Mg/k. This is also the point where the
equilibrium position and acceleration is zero.
f) The point where the acceleration is zero, s’=Mg/k.
9. Consider a two-body problem with one mass (M) on a flat frictionless table and the other (m) is
hanging vertically. Assume that M=3m.
a) Determine the speed after the hanging mass falls a distance H.
b) Explain why the height of the table doesn’t matter in this problem.
c) Redo (a) with some kinetic friction added (k). What is the
limit of the value of k?
d) Redo (c) with the table being replaced with an incline of angle ø so that M goes up
vertically (Hsinø) as m moves down H.
a) Set up 0 =∆K +∆U vfinal =[2mgh/(M+m)] 1/2=[gh/2] 1/2
b) The potential energy of the mass on the table doesn’t change.
c) Set up Wf =∆K +∆U vfinal =[(1-3)gh/2] 1/2; for v to be real <1/3.
ø
d) The work-energy principle (∆K+∆U=Wf ) here gives the following:
2
(M+m)vfinal /2 - mgH + MgHsin-MgcosH. Then you can solve for the final velocity.
H
H
10. Consider a cart of mass 40 kg moving with speed v. The cart is brought to rest over a distance of 50
cm by a pair of stiff springs of constant k1 and k2. The springs are set side by side but k2 is shorter, so
the cart compresses only k1 for 20 cm before it reaches k2. The force vs. distance graph is illustrated
below. Use the data in the graph to determine the following.
v
k2
k1
1700N
a) Determine the initial kinetic energy of the cart, and the Force on
initial speed of the cart.
cart from
springs
500N
b) The value of the constants k1 and k2.
c) The experiment is repeated with the two springs connected
.2 m
in line. How much distance does the cart move before coming to rest?
Hint: The springs compress different amounts but they exert equal forces on each other.
d) Draw the force vs. distance graph for (c). Explain the meaning of the slope in this graph?
.5 m
a) Work done to stop the cart is the area under the graph provided=380 J. This equals the change in kinetic
energy of the cart as it comes to rest. So: v= 4.36 m/s
b) k1 is the slope of the first part of the graph = 2500 N/m. The slope of the second part of the graph is (k1
+k2) = 4000 N/m; so k2= 1500 N/m.
c) Both springs are being compressed during the slowing down of the cart but they are compressed different
amounts (let’s say: x1 and x2). The kinetic energy of the cart is stored in the two springs: So, Kcart=380 J =
Uspring1 + Uspring2 = k1x12/2 + k2x22/2.
The constants are known from (b) but you also need to use the fact the springs exert equal forces on each
other, so that (k1x1 = k2x2).
You can use the force relationship to eliminate one of the variables in the energy relationship and you can
solve for the compressions that way. I got the following values: x1= 0.338 m; x2= 0.563 m.
d) The force vs. compression graph has only one slope,
844 N
the total distance that the cart moved to come to rest was
(x1 + x2= 0.90 m). The area under the graph should equal the
original K of the cart (380J). The slope of this graph represents
the combined or the equivalent spring constant ( k’= 938 N/m)
0.90 m
of the two springs acting in series. Note that in series the two
380 J
springs have a lower spring constant than one spring alone.
11. A mass m is connected to a vertical unstretched spring of constant k. The mass is released
from rest and allowed to fall until it stretches the spring the most to come to an instantaneous
stop. The unstretched position of the spring is the zero of the spring potential energy. It is
convenient in this problem to define the origin of the gravitational potential energy at the same
point, so that s=y. This makes all the gravitational potential energies negative in the problem.
For numerical values let m=0.5 kg, k= 50 N/m, and g= 10 m/s2.
s=y=0
-ymax
a) Derive an expression for the total potential energy (spring and gravitational) of the system in
terms of m, g, k, and the downward vertical displacement (-)y.
b) The total potential energy of the system is initially zero (y=0).There is another point (besides
the unstretched position) where the total potential energy is again zero. Find this other point and
explain its significance.
c) Find the point where the total potential energy is a minimum (“0” is not the minimum Utotal in
this problem) by taking the derivative of the function in (a) and setting it equal to zero (dU/dt
=0). Explain the significance of this point. Hint: What’s the acceleration at this point?
d) Fill in the following chart with the proper expressions for the different energies at the initial,
final, and middle positions. Note that in this problem we have defined y=0 where the
gravitational potential energy is the maximum. Explain how it is possible for the maximum
potential energy to equal zero.
e) Identify the location where each particular type of energy is a maximum or a minimum.
Initial position:
y=0
Middle position:
y=ymax/2
Final position:
y=ymax
Spring
Us=½ ks2
0
[min]
½ ky2 =¼ J
½kymax2 =1J
[max]
Gravitational
Ug= mg(-y)
0
[max]
-mgy=-½ J
Total U=
(ks2/2)-(mgy)
0
[max]
(ky2/2)-(mgy)
=-¼ J [min]
-mgymax=-1J 0
[min]
[max]
Kinetic K=
mv2/2
0
[min]
(mgy)- (ky2/2)
=¼ J [max]
0
[min]
Total energy
E= U + K
0 [constant]
0 [constant]
0 [constant]
b) When y=ymax, Ug+Us=0 (ky2/2) =(mgy) ymax=2mg/k=0.2 m. This represents the maximum
stretch of the spring where K=0 because the mass is turning around.
c) The minimum potential energy is found by taking the derivative of Utotal and setting it to zero:
dUtotal/dy=0. The result is y=mg/k= 0.1 m, which is the mid point between y=0 and y=max. This is the
point where the kinetic energy is the maximum and it’s also the equilibrium position of the system
(mg=kyequil) and where the acceleration=0.
d) It is possible for U=0 to be a max if the other U’s of the system add up to less than zero. This can
happen if the y=0 is chosen so that Ug< 0, as is the case here. Note that at equilibrium Utotal=-¼ J.
12. Consider a satellite of mass m orbiting a planet of mass M at a distance r from the center of the
planet. Express your answers in terms of the given quantities and G, the universal gravitational
constant.
a) Derive the expression for the total energy of the satellite in orbit.
r
b) Determine the energy that it would take to move the satellite
from being at rest on the surface of the planet to being in orbit (ignore
the daily rotation of the Earth).
c) On Earth we try to launch satellites from a location as close to the equator
as possible and toward the east. Explain what the advantage is of doing this. Hint: Consider the
daily rotation of the Earth.
d) Determine the energy that it takes to move the satellite in orbit at r to an orbit twice as far at 2r.
e) How much more energy would the satellite need to escape from its orbit at r?
f) If the satellite’s orbit decays back to the surface describe what happens to the kinetic, potential
and total energy?
This was done in class, check your notes:
a) Eorbit = Korbit +Ugrav = (GMm/2r) – (GMm/r) = -GMm/2r
b) E = (-GMm/2r) + (GMm/Rplanet)
c) The satellite has potential energy at the surface of the earth relative to infinity and it also has some
kinetic energy due to the Earth’s rotation. The closer to the equator the more kinetic energy the
satellite has before launching. This reduces the energy requirement to put the satellite in orbit.
d) E = (-GMm/4r) - (-GMm/2r) = GMm/4r
e) vescape = (2GM/r)1/2 Kescape = (GMm/r) Kescape /Korbit = (GMm/r)/(GMm/2r)= 2. This means that
it would need to double its kinetic to escape, which amounts to increasing its speed by a factor of √2.
f) Look at the formulas for these quantities: K increases, U decreases, E decreases.
13. Comets have highly elliptical orbits as compared to the planets, whose orbits are very close to
circles. Haley’s comet completes an orbit around the Sun in 75.6 years. It’s nearest position (perihelion)
from the Sun is 0.570 AU (an AU is an astronomical unit equal to the distance from the Earth to the
Sun= 1.50 x 1011 m).
a) Use Kepler’s 3rd law to show that the furthest distance that Haley travels from the Sun is about
36 AU.
b) Determine the ratio of Haley’s gravitational potential energy when nearest the Sun (perihelion)
to when it is furthest from the Sun (apehelion).
c) Kepler’s 2nd law states that the product of the velocity and the distance from the Sun is the same
for the nearest and the furthest point in an orbit. That is: vara = vprp. Determine the Haley’s
speed at apehelion if the speed at perihelion is 54.6 km/s.
d) Derive an expression for the work done by the force of gravity from the Sun when Haley’s
comet goes from apehelion to perihelion?…for a complete cycle?
e) Does the force of gravity do any work on a planet such as the Earth? Explain.
a) Use AUs and years in Kepler’s 3rd: average RHalley=(75.6)2/3 = (Rper+Rape)/2 Rape=35.2 AU.
b) Since Ug=-GMm/r Up/Ua= Ra/Rp= -35.2/-0.57.
c) From Kepler’s 2nd : vp/va= Ra/Rp va= vp (Rp/Ra)= 54.6 (0.57/35.2)= 0.884 m/s
d) Work done by gravity = Ugrav = Up-Ua=(-GMm/ Rp) - (-GMm/ Ra). For a complete cycle the work
done is zero since the force of gravity is conservative.
e) In case of a circular orbit (which Earth’s orbit approximates) the gravitational force is centripetal at
every instant and does no work.
14. A double star system is made up of two equally massive stars (M) rotating Center
of mass
about their common center of mass. The stars are a distance D apart.
a) Determine the kinetic energy of each star in terms of M, D, and G.
D
b) Determine the potential energy between the two stars.
c) Determine the total energy of the system of the two stars. Explain
why there are two kinetic energy terms in the total but only one of potential energy term.
d) How much faster would the stars have to be moving in order to escape their mutual attraction?
Similar problem to the one in the Ch. 12 homework. Remember that the gravitational force depends on
the separation between the masses (D) but each star circles the CoM of the system with radius r =D/2.
a) For each mass Fc =FG  Mv2/r=GM2/D2 v =(GM/2D)1/2 and K=GM2/4D
b) For both masses together: U = -GM2/D
c) Each mass has an amount of kinetic energy, but the potential energy really belongs to both of them
(or rather to the gravitational field between the two masses), therefore: E = K1 + K2 + U12 = GM2/2D
d) To find escape velocity compare energies at “D” to those when masses are infinitely far apart with
v=0  (Total E) infinity =0 =Kescape1 +Kescape2 +U12 =2(Mvescape2/2)-GM2/D  vescape= (GM/D)1/2.
Compare this to the speeds in orbit derived in (a), vorbit =(GM/2D)1/2 vescape / vorbit =√2
15. There are two gravitational potential energy functions to deal with in this chapter: U g= mgy for
bodies near the Earth, and Ug= -GMm/r for bodies far from the Earth. In this problem you will show
that the first expression is a simplification of the second for regions where the gravitational field is
fairly constant.
a) Write an expression for the change in potential energy of a body that is moved from r1 to a
further spot, r2, using the more general definition of Ug.
b) Simplify this expression for the case where r2 is close to r1 and are both near the surface of the
Earth.
c) Explain why the U=0 is arbitrary in the first expression, but is defined at infinity for the more
general expression.
a) In general: Ugrav = U2-U1=(-GMm/ r2) - (-GMm/ r1) = GMm(r2- r1)/ r2 r1.
b) When r2 is close to r1 (r2 ~ r1), then “g” is fairly constant from one point to the next and can be
expresses as g=GM/ r2. Also (r2 - r1)=r =h. So Ugrav = GMm(r2- r1)/ r2 r1 =mgr = mgh. And we
see how the two expressions agree with each other.
c) In the simplified expression U=mgy the location of y=0 doesn’t change the expression itself (of
course the absolute values would be different) so any arbitrary zero is equivalent to any other, but
choosing the U=0 position wisely can simplify the calculations. The grav. potential energy formula
U=(-GMm/r) is obviolusly “0” at infinity, so if we want to use that expression we must choose U=0
at infinity. Uinfinity=0 is the “wise” choice for satellites and heavenly bodies.
16. You are trying to decide whether to buy an ordinary refrigerator or a super-energy saver model. The
following chart should help you make your decision. Given the power rating and usage time of each
appliance, you can determine costs over time. Keep in mind that the power company measures energy
in a strange unit called the “kilowatt-hour” which is the energy spend in running a 1 kilowatt appliance
for one full hour. Presently, energy costs are running about12 cents per kilowatt-hour.
Appliance
Ordinary
refrigerator
Initial
cost
$800
Power rating
Daily usage
400 watts
20 % of the time
Energy Usage in
kwhs per year
80(3.15 x 107) J
=701 kW-hr
Cost over a year
~$84
Energy-saver $1200
250 watts
10 % of the time
25(3.15 x 107) J ~$26
refrigerator
=219 kW-hr
a) Determine the cost per year for each refrigerator. Use the most convenient units in this problem.
b) How many years would it take to pay back the initial investment of the energy-saver
refrigerator?
c) Do you think the investment is worth it?
There are 365x24x3600= 3.15 x 107 sec and 365x24 =8760 hrs in one year.
a) The cost/year is $84 for the ordinary refrigerator, and $26 for the energy-saver refrigerator.
This means that the energy-saver refrigerator saves $58 per year.
b) Since the energy-saver refrigerator cost $400 more to begin with, it will take (400/58)= 6.9 years to
recoup the original investment.
c) This will depend on how long you intend to keep your refrigerator. After 7 years you would be saving
money. Refrigerators usually last a long time (more than 20 years) so in most cases the initial
investment would be worth it.
17. We have seen many examples of friction doing negative work. In this problem we see friction doing
positive work to accelerate a car. A 1500 kg car accelerates uniformly from rest to 10 m/s in 3.0 s on a
level road. Neglect air resistance or other dissipative forces.
a) Find the work done on the car in this time. What force does the work
to accelerate the car?
b) Find the average power delivered to the car over the 3 s. Does your answer underestimates or
overestimates the true power output of the engine? Explain.
c) Determine the instantaneous power delivered to the engine at 2 sec.
d) Derive general expressions for the average power and for the instantaneous power for a car of
mass m that accelerates at rate a for a time t.
e) By what factor would doubling the mass of the car, without changing the driving force, affect the
instantaneous power output?
a) Wnet = K= 75 kJ. The force of static friction between wheels and road.
b) Paverage=K/t = 25 kW. This is a mininum value that doesn’t include the power needed to overcome
dissipative forces such as air resistance and friction.
c) Pinstant=dK/dt = d/dt[½ mv2] = mv(dv/dt)=mva =ma2t. Here a =3.33 m/s2, and after 2 sec v=at=6.67
m/s. So, Pinstant= 33.3 kW.
d) Paverage=K/t = K-Ko/t = ma2t/2; Pinstant=dK/dt = ma2t; so if the initial speed is zero the average
power is half the instantaneous final power value.
e) Doubling the mass would half the acceleration for the same force and half the final speed value, so
2ma’=a/2 and v’=v/2P2m =(2m)(v/2)(a/2) =mva/2=Pm/2power decreases by a factor of ½ .
18. The engine of a toy car with mass m supplies a constant power P over a time “t” to the wheels to
accelerate the car. You can ignore air resistance and other dissipative forces. Assume that the car starts
from rest.
a)
b)
c)
d)
Show that the speed of the car as a function of time is: v= (2Pt/m)1/2.
Show that the acceleration is not constant in this case.
Does the expression in (b) make sense for t=0? Justify your answer.
If your 1500 kg car accelerates from zero to 60 mph (about 27 m/s) in 5 seconds. What is the
average power of your car? Find this answer in both kilowatts and in horsepower.
a) K=(P)(t) = mv2/2 v=(2Pt/m)1/2.
b) a=dv/dt= =(P/2m)1/2(t)-1/2.
c) The expression in (b) doesn’t make sense at t=0 because it predicts an infinite acceleration which is not
possible. This means that it is really impossible to maintain a constant power from t=0.
d) Paverage=DK/Dt = 109 kW = 146 hp.
19. Ignoring air resistance is a large error when dealing with cars, especially when they are traveling at
high speeds. Remember that air resistance depends on the speed squared: FAR=-Av2.
Problem 19
Problem 20
a) Draw a “FBD” for the car. Show that the power required to keep a car going at constant speed
on a flat road against air resistance alone is proportional to the speed to the third power.
b) If your car consumes 20 hp when traveling at a constant speed of 30 mph. How much more
power would it need to travel at a constant speed of 60 mph under the same conditions?
c) What are the units of the constant “A” in the mks system? Remember that the constant A varies
for different cars.
d) Explain why cars are generally less efficient at higher speed than at lower speeds.
This was done in class;
a) P=Fv for a constant force. Here F=Av2, so P =Av3.
b) P2/P1 =(v2/v1)3 = 8/1.
c) Units of “A” are Ns2/m2.
d) Cars are generally less efficient at higher speeds because of the results in (b). Even if you consider
that you save half the time by doubling the speed, the 8-fold increase in power means that you would
spend 4 times more energy to travel the same distance at twice the speed. That is, of course, if we
only consider air resistance. There are other dissipative forces such as rolling friction and the
internal workings of the engine that would also generate heat rather than motion.
20. The same car as in the problem above now finds itself going up an incline of angle ø.
a) If the car want to maintain the same speed vo as on the flat road, how much more power does it
need?
b) If the car engine in (a) cannot increase its power it must slow down when going uphill. If the car
is forced to slow down to half its speed on the incline, determine an expression for the angle of
the incline.
c) At what angle would the car coast downhill at the speed vo?
a) Pincline =Pair res + v(mg sinø). This last term is the extra power needed.
2
b) Pincline =Pflat  A(vo/2)3 +( vo/2)mg sinø = A(vo)3  sinø
o /mg)(14/8)
2
c) Here Pair res = vo(mg sinø) sinø= (Avo /mg)
21. The graph shown illustrates the potential
energy associated with a conservative force.
U
b
a
c
d
x
a) Identify the points of stable equilibrium, unstable equilibrium, and neutral equilibrium (if any).
b) If a particle is located between points b and c, will it be pushed in the +x direction or the –x
direction? Justify your answer.
c) If the potential energy function is: U(x) = A e--Bx cos (Cx) ; where A,B and C are constants.
Determine the force function.
d) Make a rough sketch of the force function vs. distance. Hint: Don’t try to plot the function
derived in (c), merely take your clues from the slopes of the U graph provided.
a) b is stable; 0 and d are unstable; infinity is neutral
b) Between b and c dU/dx is (+) so F is (-); which means the particle is pushed back to the
equilibrium point at b.
F
c) F =-dU/dx =Ae-Bx (BcosCx + CsinCx)
d) Something like this 
d
b
c
a
x
22. Draw a graph of the potential energy function for the vertical spring and mass in problem 11.
a)
b)
c)
d)
e)
Identify the point(s) of equilibrium on the graph.
Identify the point(s) where the kinetic energy is a maximum.
Identify the point(s) where the spring potential energy is minimum….is maximum.
Identify the point(s) where the gravitational potential energy is minimum…is maximum.
Identify the point(s) where the total potential energy is minimum…is maximum.
bottom
Note that all positions are negative by definition of y=0.
ymin=-2mg/k
a) Stable equilibrium at y=-(mg/k).
b) Max K at the point of equilibrium y=-(mg/k).
c) Spring U is a min at y=0; and it is a max at y=-(2mg/k).
d) Gravitational U is a max at y=0; and it is a min at y=-(2mg/k).
e) Total U is max at y=0 and at y=-(2mg/k); it is a min at yequil=-(mg/k).
s=y=0
-ymax
equilibrium
Utotal
yeq=-mg/k
yo =0
Umim= -(mg)2/2k
Challenge problems:
23. The following problem will help you understand why the force of air resistance depends on the
speed squared. It will also show you how scientists use models to develop a theory. Let us model a car
as a simple cylinder of length L and cross-section Across that is moving with speed v. Assume the
surrounding air has a density .
vt
Volume of air
displaced =A crossvt
v
y
Across
a) In order to move forward the car must do work on the volume of air in front of it and accelerate
it up to its own speed v. Use this idea to derive an expression for the mass of air (m) that is
accelerated by the cylinder in a time (t).
b) Determine an expression for the kinetic energy acquired by the air mass (m) in the time (t).
c) Since power can also be expressed as the energy transferred per time, derive the expression for
the power generated by the car while moving forward at constant speed.
d) Use the power expression to determine the force of air resistance in this model.
e) What is the air resistance constant in this model?
f) This model is obviously oversimplified. What other factors do you think should be considered
in deriving a more general expression for air resistance?
a) m= Density x volume= Acrossvt)
b) K=mv2/2= Acrossv3t /2
c) K/t = Acrossv3/2
d) Here P=FARv  FAR=P/v=Acrossv2/2 and we see that the force or air resistance is proportional to
the speed squared of the moving object.
e) The constant we often label “A” in the formula for air resistance is equivalent in this model to
“A”=Across/2
f) Other factors could be the shape of the object, and the viscosity of the air, as well as turbulence of
the air as it moves around the car.
24. For the future mechanical engineers…Pulleys and incline planes are known as “simple machines”
that help us raise heavier objects than we could lift directly. However, the advantage that we get from
these machines is merely a trade-off between force and distance. Consider using a double pulley system
and also a ramp (with and without friction) in order to move a load of mass M up a vertical height H at
constant speed.
Engineers define the “real mechanical advantage” (RMA) as the ratio of the load lifted to the force
used to lift it. They also define the “ideal mechanical advantage” (IMA) as the ratio of the
displacement of the point of the applied force to the displacement of the load.
U of M
2/1
Work done by
pulling force
MgH
H/sin
1/sin
MgH
MgH
1/(sincos H/sin
1/sin
MgH(1 cot
MgH
Illustration
Pulling Force
RMA
IMA
2/1
Pulling
distance
2H
a)
Mg/2
Mgsin
1/sin
Mg(sincos
MgH
H
Puller
M
b) 
=0
F
H

c)  > 0
F
H

d) Are your results in (a) and (b) consistent with energy conservation? Explain.
e) Is the “RMA” always equal the “IMA”? Explain.
f) In a previous problem, you proved that the minimum force needed to push a body up a ramp at
constant speed against friction occurs when tan=1/. In pushing up the block, determine the ramp
angle for which: (i) the RMA is a maximum; (ii) the IMA is a maximum; (iii) the work done is a
minimum.
d) As can be seen in (c) RMA is sometimes (actually often) less than IMA. That’s because we often have
to work against forces like friction that dissipate energy as we try to do useful work.
e) (i) arctan0; 0o.