Section 1-6 - Work & Energy

Work & Energy

How was that quality transferred into the object?  A force was applied, but the force must have acted through a displacement.  Let us call the transfer of energy into (or out of) an object the work done on the object.  The bigger the force, the more energy is transferred, and the greater the displacement over which the force acted, the more energy is transferred.  What's more, there is an effect due to the relative orientation of the force with the displacement:
If F and Dx are in the same direction, energy is transferred into the object and we say that positive work was done.
If F and Dx are in the opposite directions, energy is transferred out of the object and we say that negative work was done.
If F and Dx are perpendicular, no energy is transferred into the object and we say that no work was done.
How can we write this in a more mathematical way?  Let the work W be defined as:
W = F (Dx) cosq_F,Dx,
that is, as the magnitude of the force times the magnitude of the displacement, times the cosine of the angle between  those two vectors.  So,
if F and Dx are parallel, q_F,Dx= 0^o and W = FDx (positive).
if F and Dx are anti-parallel, q_F,Dx = 180^o and W = - FDx (negative).
if F and Dx are perpendicular, q_F,Dx = 90^o and there is no work done.

The guess of the cosine function is not quite as arbitrary as it seems.  Consider a force applied at some random angle as shown here.

We can always decompose the force vector into a component perpendicular to the displacement (which will do no work) and one parallel (or anti-parallel) to the displacement which will perform work
W = (Fcosq) Dx,
as before.  This leads us to an alternate way of writing the definition of work:
W = F_|| Dx.
Indeed, we could also look at the work as the whole force times the component of the displacement in the direction of the force:
W = F Dx _||.

It might be useful to review vector multiplicaton at this point.  There are several types of multiplication, of which we will discuss two. We define the scalar product (also called the inner product or the dot product) of two vectors A and B to be: A.B = |A| |B| cosqA,B, = ABcosqA,B, that is, the magnitude of A times the magnitude of B times the cosine of the angle between them.  One interpretation of this definition is that we are multiplying the magnitude of A vector by the component of B that lies in the direction of A: A.B = AB|| = ABcosqA,B. The work is an example.  We can write that W = F.Dx. Another type of vector multiplication is the vector product or the cross product: AxB.  We define the magnitude of the cross product to be |AxB| = |A| |B| sinqA,B, = AB sinqA,B. The direction of AxB is perpendicular to the plane that contains A and B and can be obtained by using the right-hand-rule (RHR).  Point the index finger of the right hand in the direction of A and the middle finger in the direction of B; the right thumb then points in the direction of the cross product.  One interpretation of the cross product's magnitude is that it is the area of the parallelogram formed by the vectors A and B when they are placed tail to tail: The base of the parallelogram is A and the height is Bsinq, making the area A(Bsinq).  What is the direction of AxB in this example? Answer Perpendicularly out of the page toward you. We'll discuss quantities that can be written as cross products later.

What if the force applied were not constant (or in other words, we have a variable force)?  Clearly, more work would be done in some displacement intervals than in others.

We need to break the overall displacement down into very small displacements Dx_n, over which we can consider the force to be relatively constant at value F_Dxn; we then find the work done over that interval to be W_Dxn  = F_Dxn Dx_n .  Then the work done by the force is
W_F = S_DxnW_Dxn = S_Dxn F_Dxn Dx_n
This value approaches the area under the force-displacement curve as the widths of the Dx_i's go toward zero.

W =  F ^.dx.
What if the curve were to go below the axis?  Then the work would be negative.  What if the Dx values were negative?  That would also switch the sign of the work.
What if we're working in 3d?  Then we write that
W = F(r) ^.dr,
which gives one term in each of the x, y, and z directions.

Let's stop for a second.  We've defined the work, but we can define anything we please to be whatever we please; what's the point?  The definition is meaningful only if it is useful.  Let's think back to the beginning of this discussion.  We talked about the work on an object as a transfer of energy, so that we should be able to say that the work is the change in the amount of this energy stuff that the object possesses, i.e., W = DE.  Keep this in mind as we do a little derivation:

Consider the two relations, where the acceleration is not constant:

a = dv/dt  and  v = dr/dt

This time, let's perform a dot product on each side to obtain

a ^. dr/dt = v ^. dv/dt

a ^. dr = v ^. dv

We might assume that the acceleration is a functiom of time, but it might as well be a function of position instead.

a(r) ^. dr = v ^. dv

Let's make use of Newton's second law, F_NET = ma, to obtain

(1/m) F_NET(r) ^. dr = v ^. dv

and integrate

_r1^r2 F

_vi^vf

Note that it is nothing more than Newton's Second Law, combined with two definitions.  Why bother?  We will find that this picture will be on occasion more convenient to use than forces and accelerations, especially in cases where we don't need to know the time a trip takes, or when the acceleration is not constant (see Note 1 below).

Note also that work is a scalar.  Since the kinetic energy depends on the speed of an object, and not the direction of the object's motion, it is a scalar; the work is the change in the KE, so it too is a scalar.  Although we did this derivation for one dimension, it can be done quite easily for three with the same result (see Note 2 below).  A combination of these arguments let's us assert that the result is valid even for variable forces in three dimensions.

Consider a block of mass m = 5 kg at the top of a frictionless ramp L = 2 metres long, which is inclined at q = 37^o to the horizontal.

If the mass starts from rest at the top, how quickly will it be moving when it reaches the bottom?  Draw a free-body diagram:

W_N = 0, since the force is perpendicular to the displacement
W_g = mg L cosq_mg,Dx
W_N + W_g = ¹/₂mv_f² - ¹/₂mv_i²
mg L cosq_mg,Dx = ¹/₂mv_f²
v_f² = 2g L cosq_mg,Dx
v_f = [2g L cosq_mg,Dx ]^1/2 = [2*10*2*cos53^o]^1/2 = 4.9 m/s.

Now, let's suppose that the block started out with an initial velocity of 6 m/s down the plane.  Make a guess what the speed will be at the bottom:

Are you surprised?  What went wrong with your guess?  Isn't the acceleration the same in each case?

Here is another throught question:  Suppose that I drop an object from a given height; the force of gravity (the object's weight) does work and the kinetic energy of the object increases.  Now, suppose instead that I slowly lower the object slowly from the same initial altitude.  Compare the work done by gravity in the second case to the work done in the first case.

Conservative and non-Conservative Forces

There are a number of ways to define what a conservative force is.  I like to say that a conservative force is one for which the work it does on an object moving from Point A to Point B is independent of the path the object takes.  Let's take the weight of an object as a concrete example.  Suppose that I lower a mass m from a height h above the table to the top of the table.  I'm only interested at this point in what the weight does, not what any other force, such as from my hand, does.  The force is mg downward, and the displacement is h downward, and those two vectors are parallel, so we have that

W_g = (mg)(h)(cos0^o) = mgh.
Now, let's take the object on a little tour of the region.  Move it horizontally a displacement s, then down h, then horizontally again s', back to point B:

The work done will be
W_AB = mg s cos 90^o + mg h cos 0^o + mg s' cos 90^o = mgh,
again.
Let's pick a random path:

You might be able to see that we can always approximate any path to an arbitrary degree of accuracy with these stepped horizontal and vertical movements.  From previous discussion, we know that any horizontal movements will correspond to no work being done by gravity.  The vertical displacements are each of magnitude h_i , some parallel to the weight and some anti-parallel, such that the work done by the weight during each vertical motion is
W_vertical = S_nmg h_n cosq_n = mg S_n h_ncosq_n, where cosq_n = +1 if the displacement is downward (parallel to the force) and -1 if the displacement is upward (anti-parallel to the force).
We realize that S_n h_ncosq_n  = h,
and see that
W_vertical = S_n mg h_n cosq_n = mgh,
as before, so that the work done by the weight throughout the whole trip is
W_AB = mgh,
independent of the path taken.

Let's consider an example of a non-conservative force: friction.  Consider an object being slid across a table top along two paths (let all Dx's be the same magnitude):

Remember that we are not concerned with the work done by any other force, such as that of the hand which pushes the block.
The frictional force will be (not proven here):
F_f = m_Kmg,
so that the work done by friction from Point A to Point B along the direct path is
W_direct = F_f Dx cos(180^o) = -m_Kmg Dx.
Along the indirect path, this will be three times bigger:
W_indirect =  -m_Kmg Dx₁ - m_Kmg Dx₂ - m_Kmg Dx₃. = -3m_Kmg Dx.
So, we see that friction is not a conservative force.

Potential Energy

To do this, we divide the total work on an object into two types, depending on whether each force is conservative or non-conservative:
W_cons + W_non-cons = DKE
- DPE + W_non-cons = DKE
W_non-cons = DKE + DPE.
Remember that there may well be more than one conservative force operating on the object, which would require us to have more than one DPE term.  Also remember that one should not put the term on both sides of the relationship; don't count gravity's effect as both a work and as a potential energy.

Can we figure out what the gravitational potential energy function is?  Not really.  We can only figure out an expression for the change in the PE.  Suppose that we let y be the vertical location of the object, and that we lift (or lower) the object by some displacement Dy.  The force of gravity is of course downward, so if we lift the object, Dy is positive, the angle between the weight and the displacement is 180^o, so the work done is equal to (mg)(Dy) (-1) = -mg Dy, while the change in PE is the negative of that, or +mgDy.  If we lowered the object, the signs of each term (work and DPE) would correspondingly reverse.  So
DPE_grav = mg Dy,
and it seems O.K. to say that
PE_grav = mgy,
so long as we keep in mind that the assignation of PE = 0 is arbitrary.

Later, we'll discuss  the potential energy from a different conservative force.

As an extra comment, consider the following:

DPE = -W
d(PE) = -F(x) dx
Then,
d(PE) = -F(x) dx
F(x) = - d(PE)/dx

Conservation of Mechanical Energy

For example, let's look once again at the falling pen.  Just after release, the pen has zero KE and mgh of PE (we'll let PE = 0 at the table top).  Just before hitting the table, the PE = 0 and the KE is not zero, and in fact equals numerically mgh:
KE_i +  PE_i = KE_f  + PE_f
0 + mgh = ¹/₂mv_f² + 0.

Right after the pen hits the table, it has no PE and no KE!  What happened to the energy?

Springs

Is the spring force conservative?  A quick look suggests that it is.  Since the force exerted by the spring depends only on the position of the end of the spring (we'll assume that the other end is fixed), reversing the displacement back over already covered ground simply undoes the work done the first time (by flipping the sign of the cosine term), so that the net work done depends only on the initial and final positions of the end of the spring.

How much work is necessary to stretch (or compress) a spring distance x from its relaxed position?  We can use the graphical representation showing F_{on spring} as a linear function of c (slope = k):

We showed previously that the work done by any variable force is represented by the area under the curve.  Since this is a triangle, the area is one-half the base times the height:
A = ¹/₂bh = ¹/₂(c) (F) = ¹/₂(c)(kc) = ¹/₂kc².
So, the work done on the spring is ¹/₂kc², the work done by the spring is -¹/₂kc², and the change in the PE of the spring is the negative of that, or
DPE = +¹/₂kc².
If we define the PE to be zero at c = 0 (the relaxed position), then we can say more simply that PE_spring = ¹/₂kc².

Let's do this again using calculus.  Move the spring from its relaxed position to some arbitrary position, c.
W_{on spring} = ₀ ^c F_spring(c) dc = ₀ ^c kc dc = ¹/₂kc² - 0.
The work done by the spring will be, by the third law,
W_{by spring} = - ¹/₂kc².
Then, since DPE_sp = - W by spring,
PE_{sp f} - PE_{sp i} = + ¹/₂kc².
Since we can choose the PE to be zero wherever we please, it seems natural to define PE_sp = PE_{sp i} = 0 when the spring is relaxed, so that
PE_sp = ¹/₂kc².

Now, some fun with springs.  Occasionally, we see more than one spring attached to an object.  What would be the effective spring constant of more than one spring, i.e., what would k_EFF for a single spring be to do the same job as the multiple springs?
If the springs are both attached to the object, it's easy:

FIGURE
F_TOT = F₁ + F₂ = -k₁c₁ - k₂c₂ = -(k₁ + k₂) c = -k_EFF c;   k_EFF = k₁ + k₂.
If the springs are attached end to end,
FIGURE
F_TOT = F₁ = F₂
c₁ + c₂ = c_TOT
F₁/k₁ + F₂/k₂ = F_TOT/k_EFF
1/k_EFF = 1/k₁ + 1/k₂

(PE) = -W = -F(x) dx,  then

d

(PE) = -F(x) dx,  and then
d

Power

We obtain an interesting result if we re-introduce the definition of the velocity, v = dx/dt:
P = dW/dt = [F dx cosq]/dt = Fcosq[dx/dt] = Fvcosq.

Examples:

Start with the Work-Energy Theorem:
W_NC = DKE + DPE
What forces act on the block?  There's the weight, the normal force from the plane, friction, and the spring force.  The weight and spring force can be treated as potential; energy terms, and the normal force does no work (it's perpendicular to the displacement). The friction acts up the plane as the mass moves down the plane (i.e., cosq_F,Dx = -1). Use NII to find F_f.  Let perpendicular/upward be positive y:
N - mgcosq = ma_y = 0
F_f = m_KN = m_Kmgcosq
Then,
-m_Kmgcosq Dx = KE_f - KE_i + PE_gf - PE_gi + PE_Sf - PE_Si
Since the mass starts and ends at rest, both KE terms are zero.  The spring starts out unstretched, so PE_So = 0,  Let y = 0 where the mass comes to rest, so that PE_gf = 0.
-m_Kmgcosq Dx = - PE_gi + PE_Sf = -mgy_i + ¹/₂k[c]²
We need to relate the vertical height y_i to the distance down the plane Dx:  y_i = Dx sinq.
-m_Kmgcosq Dx = -mgDx sinq + ¹/₂k[c]²
m_K = [mgDx sinq - ¹/₂k[c]²]/mgcosq Dx = [sinq - ¹/₂[k/mg]c]/cosq = tanq - kDx/2mgcosq
m_K = tan37^o - 100(0.2)/2*2*9.8*cos37^o = 0.11

Another Example:
Consider a massless spring of constant k hanging from the ceiling.  Let's attach a mass m and allow the mass to settle very slowly to an equilibrium point.
A) How far does the spring stretch?  What is the total energy in this situation?
Now, let's raise the mass back to the spring's relaxed position and drop it.
B) How quickly is the mass moving as it passes through the equilibrium point?
C) How far will the mass drop before stopping?

Let's let where the spring is relaxed be y = 0 and let y be positive upward.  When the mass is first attached, the KE is zero (no motion), the PE_S is zero (the spring is relaxed) and the PE_g is zero (because we said so), so the total mechanical energy is zero.  Once we lower the mass to its equilibrium position and let go, we know that the net force on it is zero:
-mg - kc = 0,
so c = Dy = y_f - y_i = y_f = -mg/k
It makes sense that this is a negative number, since the mass would certainly have descended.  Since the mass is at rest, there is no KE, and the PE_S will be
PE_S = ¹/₂k[c]² = ¹/₂k[y_f]² = m²g²/2k
and the PE_g will be
PE_g = mgy_f = - m²g²/k,
meaning that the total energy is
E_TOT = -m²g²/2k < 0!
How can this be?

Now, return the mass back to the spring's relaxed position and drop it.  What is the total energy as it passed through the equilibrium point?
Well, we can write an expression, but can't actually calculate it:
E_TOT = PE_S + PE_g + KE
Now, the PEs we did above: m²g²/2k + -m²g²/k  = -m²g²/2k
Note that we do expect E_TOT to be zero, since this time there were no non-conservative forces, and mechanical energy should have been conserved. So th eKE should by given by
KE = E_TOT - PE_TOT = 0 - -m²g²/2k = +m²g²/2k.
From this, we can find the mass's speed as it passes through the equilibrium point:
v = [m/k]^1/2g.

Eventually, the mass will stop (KE = 0) and then start moving back upward.  Where does this happen (y_ff)?
E_TOT = 0 = KE_ff + PE_Sff + PE_gff = 0 + ¹/₂k[y_ff]² + mgy_ff
Note that there are two solutions to this: y_ff = 0 (i.e., at the top) and y_ff = -2mg/k.

Another Example:
A toy gun launches its projectile by means of a spring, with unknown spring constant k.  If the spring is compressed 0.12 m from its relaxed position and fired vertically, the gun can launch a 20g projectile from rest to a height of 20m above its initial position.  Find the spring constant k and the speed of the projectile as it passes through the spring's equilibrium position  (This is S&F 5-28):

Assuming no friction, we can say that there are no non-conservative forces doing work.  So,
W_NC = 0 = DKE + DPE_GRAVITY + DPE_SPRING
0 = ¹/₂mv_f² - ¹/₂mv_i² + mgy_f - mgy_i + ¹/₂kc_f² - ¹/₂kc_i²
Here, c represents the amount the spring is stretched or compressed and y is the altitude of the ball.
The object starts at rest and ends at rest, so v_i = v_f = 0.
Let the vertical reference level be where the spring is relaxed, so y_i = c_i = - 0.12 m, y_f = 20m, and c_f = 0 (the spring is again relaxed):
0 = mgy_f - mgy_i - ¹/₂ky_i^2
1/₂ky_i² = mgy_f - mgy_i
k =  2mg[y_f - y_i]/y_i² = 2*0.02*9.8[20 - ^- 0.12]/(0.12²) = 547 N/m
Now, go back and find the speed of the object as it passes the spring's relaxation point.  Use the same basic relationship:
0 = ¹/₂mv_f² - ¹/₂mv_i² + mgy_f - mgy_i + ¹/₂kc_f² - ¹/₂kc_i²
but now, v_i = 0, y_f = x_f = 0, and c_i = y_i = -0.12 m.
0 = ¹/₂mv_f² + - mgy_i - ¹/₂kc_i^2
1/₂mv_f² = mgy_i + ¹/₂kc_i²
v_f² = 2gy_i + (k/m)c_io²
v_f = [2gy_i + (k/m)c_i²]^1/2 = [2*9.8*(-.12) + (547/0.02m)(-0.12)²]^1/2 = 19.3 m/s

Yet Another Example:

Consider a skier at the top of a hemispherical knoll of radius R, which is covered in slippery snow.  He starts from rest at the top and travels down the side.  At what vertical distance h from the ground will he become airborne?

What forces act on the skier?  Are any non-conservative forces doing work?

Note 1:

Note 2: