Answers/Solutions: Giancoli Chapter 8

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Q3: If the net force is conservative, then the total energy (KE + PE) must be conserved. Thus if KE increases by 300 J, the PE must decrease by this same amount. Thus \Delta \ PE = - 300 J. Comment: Since all forces are conservative, total mechanical energy (KE+PE) energy is conserved i.e., will not change.

Q5: Sliding down the hill with no friction, only two forces act on the sled: the normal force (which does no work) and gravity (which acts directly downward). When the sled slides downward through a given height, regardless of the angle of the incline, the distance moved in the direction of the gravitational force is the same, and so the same amount of work is done. So the final KE does not depend on the steepness.

If, on the other hand, friction acts, the frictional force does negative work on the sled all along the length of its path. In this case, the amount of this work depends on how far along the slope the sled moves. So the net work done (and thus the final KE) does depend on the steepness.

Q15: Yes. Clearly the KE cannot be negative, but PE can easily be negative, depending on where one sets the arbitrary zero-level. If the PE is, say - 500 Joules while the KE is +200 Joules, then the total mechanical energy would be - 300 Joules.

Q26: As seen in Equation 8-7 on page 189, the force is the negative derivative of the PE:

F(x) = - dU/dx

(a) The maximum force occurs where the slope is steepest. I judge this to be point C.

(b) The sign of the force is opposite that of the slope. Just imagine a ball rolling on the curve and consider which direction it "wants" to roll. The force would be to the left at points A, E, and F, and to the right at point C.

(c) Equilibrium occurs where the slope is zero: at points B, D, and G. (Admittedly, the black dot at G is a little misplaced: it is easy to think it is to the right of the peak, which would mean the force would be in the positive direction there. But I think the author intended the G to be at the peak.) B is a point of "neutral" equilibrium; D is a point of "stable" equilibrium; and G is a point of "unstable" equilibrium.

P3: Two vertical forces act on the mass: gravity downward and the spring force upward. At equilibrium, these must balance, so if \Delta \ y is the "stretch" of the spring:

k\Delta \ y = Mg

which yields \Delta \ y = 0.39 m = 39 cm. If the original position was 15 cm, the final position (assuming the meter stick was oriented so that the reading increased downward) would be 54 cm.

P8: The force in the x-direction is the partial derivative of the PE with respect to x: Fx = \partial U\ /\partial x\ .

The force in the y-direction is the partial derivative of the PE with respect to y: Fy = \partial U\ /\partial y\ .

The force in the z-direction is the partial derivative of the PE with respect to z: Fz = \partial U\ /\partial z\ .

(Taking a partial derivative is easy: when taking the derivative w.r.t. x, you simply treat the y and z as constants.)


Fx = 6x + 2y

Fy = 2x + 8yz

Fz = 4y2

P18: (a) Since there are no horizontal forces acting on the ball, and since its velocity is horizontal at its highest point, its velocity there will simply be its original x-velocity:

vx = vocos(\theta \ o) = (8.5)cos(36) = 6.88 m/s

(b) Total energy is conserved, so the ball's gain in PE (= Mg\Delta \ h) at its highest point is equal to the difference between the original KE and the KE at the highest point. So:

Mg\Delta \ y = (1/2)Mvo2 - (1/2)Mvt2

where vt is the velocity at the "top" of the path from part (a). The mass divides out, and the result is \Delta \ y = 1.27 m.

P20: Let's set the zero for gravitational PE at position 2. Then the PE at point 1 is Mg(32) and the KE is zero at this point.

At point 2: Conservation of Energy gives:

(KE)2 + (PE)2 = (KE)1 + (PE)1


(1/2)Mv22 + 0 = (1/2)M(0)2 + Mg(32)

The mass divides out and v2 = 25.0 m/s

At point 3: (KE)3 + (PE)3 = (KE)1 + (PE)1

(Comment: we could have used point 2 instead of point 1 here), so

(1/2)Mv32 + Mg(26) = (1/2)M(0)2 + Mg(32)

Again, the mass divides out and v3 = 10.8 m/s

At point 4: (KE)4 + (PE)4 = (KE)1 + (PE)1

(Comment: we could have used either point 3 or point 2 instead of point 1 here), so

(1/2)Mv42 + Mg(14) = (1/2)M(0)2 + Mg(32)

and the result is v4 = 18.8 m/s

P28: Comment: One of my all-time favorite problems. It is not suitable for a Test, but maybe for the Final Exam! The reason it's not suitable for a Test - in addition to being difficult - is that it requires use of knowledge from several earlier chapters which were covered on earlier tests. But this is precisely why it would be a great final exam problem!

(a) Clearly we are going to use the principle of Conservation of Energy (That's the Chapter title!). Draw a free-body diagram of the skier after she has moved through an angle \theta \ (as in Figure 8-36). Only two forces act: gravity (straight down) and the normal force (along a radius, perpendicular to the sphere's surface). But just as the skier leaves the sphere's surface, this normal force is zero. Of course that is the moment when the skier's path is no longer circular. SO: the condition for the skier leaving the surface of the sphere is that the component of the weight along the radius provides just exactly the needed centripetal force to keep the skier moving in a circle at speed v and radius r. From your free-body diagram:

(mg)cos(\theta \ ) = mv2/r

Of course we don't yet know what v is. Well, let's use energy conservation to figure out v. When the skier is at angle \theta \ she has descended a distance [r - (r)cos(\theta \ )] from the top of the hill. Thus an amount of PE equal to (mgr)(1 - cos\theta \ ) has turned into KE:

mgr[1 - cos(\theta \ )] = (1/2)mv2

The mass m and the radius are "given," so they might appear in the answer. BUT the mass divides out of both equations, so it will not, in fact, appear in the answer. Note that we now have two equations with just two unknowns: v and \theta \ . We are looking for \theta \ , so we eliminate v: in fact, after dividing out m, all we have to do is equate the two different expressions for v2:

(gr)cos(\theta \ ) = (2gr)[1 - cos(\theta \ )]

At this point both g and r divide out, and we can trivially solve for the cosine, and then the angle:

cos(\theta \ ) = 2/3

which gives

\theta \ = 48 degrees.

Comment: It is fascinating that this answer does NOT depend on either m or r: it will be the same for ANY object sliding on a smooth (no friction) sphere!

(b) I'm going to leave it to you to figure out how friction, if present, would affect the answer for \theta \ . You are not asked to find a number here, only whether the angle would be greater or smaller than the answer in part (a).

P36: Pre-Comment: I love this problem, and there will almost certainly be one like it on the Test. I'm going to set it up and get the answer for part (d), leaving you to get the answers for the other parts.

Looking at Figure 8-37, it is pretty easy to "see" what is going to happen: the block will slide down the curved part of the track, gaining speed until it reaches point B. From this point until point C it will lose speed as friction slows it down. At point C it will hit the spring, compressing it some distance (call it \Delta \ s) before momentarily coming to a stop. We are asked (in part d) to compute the spring constant. We are given the mass m, the radius r, the distance over which there is friction (call this \Delta \ x), the coefficient of kinetic friction over this distance, and the maximum compression of the spring \Delta \ s.

Obviously we are going to use conservation of energy (it's the Chapter title). All forces acting except the frictional force are conservative. We will use a form of Equation 8-14 on page 196, combined with the fact (see top of p. 197) that a frictional force acting over a distance \Delta \ x dissipates an amount of energy equal to Ffr\Delta \ x. In fact, the equation we need is written - without a number - on page 197; in my notation:

\Delta \ KE + \Delta \ PE + Ffr\Delta \ x = 0

So: setting the zero for gravitational PE at the bottom of the track, and the zero for spring PE at the relaxed position of the spring, the energy at various points looks like:

EA = mgr

EB = (1/2)mvB2

EC = (1/2)mvC2

Ef = (1/2)k(\Delta \ s)2

Now if there were no friction, each of these terms would equal all the others. But friction dissipates some of the mechanical energy between points B and C, so the energies at C and f (which are equal) is less than that at A and B (which are also equal) by an amount

Elost = Ffr\Delta \ x

The answers for all parts of this problem are obtained by equating various combinations of the above terms. For part (d), since we seek k, which appears in Ef, and since we know everything in EA and in Elost, we write

Ef - EA + Elost = 0

which is just our form of Equation 8-14 (make sure you understand this - play with the two equations, writing out each term - until you are confident they are the same!).

We have, then:

(1/2)k(\Delta \ s)2 = mgr - Ffr\Delta \ x

which we need to solve for k, resulting in

k = 2mg[r - \mu \ k\Delta \ x]/(\Delta \ s)2 = 612.5 N/m

Post-Comment: I've made this explanation much too long - it's my greatest failing as a teacher: I am desperate to be "clear" and therefore am too "wordy." The problem should be pretty easy: The initial energy (which is gravitational PE) all gets converted to KE. Then part of this is lost to friction, resulting in a smaller KE. Finally, the remaining KE is converted to spring PE. We have simple expressions for each of these amounts of energy, and by keeping careful track of the energy, we can set up an equation that will let us solve for the sought-for quantity. If you can keep track of your checking account, this problem should not be hard.

P62: The amount of energy needed to raise a mass M a height h is Mgh. If the motor's power (Energy/time; the rate at which the motor provides energy) is P, then the time required to generate Mgh is

time = Energy/Power = Mgh/P = (335 kg)(9.8 m/s2)(16.0 m)/(1750 Joules/sec) = 30 seconds

P70: The motor's energy output in one minute is mgh = (21 kg)(9.8 m/s2)(3.5 m) = 720.3 Joules. Thus the rate at which energy is output is 12.0 Joules/second, or 12.0 Watts.

P73: We have the mass M and the position as a function of time x(t) and this allows us to know just about everything! For example, the instantaneous velocity (i.e., the velocity at any instant) is:

v(t) = dx/dt

and the acceleration at any instant is:

a(t) = dv/dt = d2x/dt2

Then the net force on the object must be:

F(t) = ma(t)

Finally, the power at any instant will be:

P(t) = F(t)\bullet \ v(t)

as in Equation 8-21. The KE at any instant will be:

KE = (1/2)Mv2,

so the average power over any time interval \Delta \ t will be:

PAV = \Delta \ KE/\Delta \ t

Now let's USE these equations! We have M = 280 grams ( = 0.28 kg), so:

x(t) = 5t3 - 8t2 - 44t

v(t) = 15t2 - 16t - 44

a(t) = 30t - 16

Finally, answering the specific questions:

(a) P(2) = F(2)v(2) = Ma(2)v(2) = (0.28)(44)(- 16) = - 197 Watts

(b) P(4) = F(4)v(4) = Ma(4)v(4) = (0.28)(104)(132) = +3844 Watts

(c) PAV = (KEf - KEi)/(tf - ti)

so, between t = 0 and t = 2,

PAV = [(1/2)(0.28)(- 16)2 - (1/2)(0.28)(- 44)2]/2 = - 117.6 Watts

and, between t = 2 and t = 4,

PAV = [(1/2)(0.28)(132)2 - (1/2)(0.28)(- 16)2]/2 = 1202 Watts

Comment: The meta-lesson here is that once we know the position of a particle as a function of time, we know an awful lot about what is happening and will happen to the particle!