If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

015.  `query 15

Question:  `qSet 4 probs 1-7

If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?

Your solution: 

Confidence rating::

Given Solution: 

`a** You can find the change in the momentum.  Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **

Self-critique (if necessary):

Self-critique rating:

Question:  `qWhat is the definition of the momentum of an object?

Your solution: 

Confidence rating::

Given Solution: 

`a** momentum = mass * velocity.

Change in momentum is mass * change in velocity (assuming constant mass).

UNIVERSITY PHYSICS NOTE:  If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm.  If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **

STUDENT QUESTION

ok, I am confused what 'dp stands for in your explanation.

INSTRUCTOR RESPONSE

p is the standard symbol for momentum.

Thus `dp is the change in momentum. `dp = `d( m v ).

Self-critique (if necessary):

Self-critique rating:

Question:  `qHow do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?

Your solution: 

Confidence rating::

Given Solution: 

`a** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum.  **

Self-critique (if necessary):

Self-critique rating:

Question:  `qHow is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?

Your solution: 

Confidence rating::

Given Solution: 

`a**  First from F=ma we understand that a=F/m.

Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. 

Since vf-v0 = 'dv, this becomes 'dv = a'dt.

Now substituting a=F/m , we get

'dv = (F/m)'dt    Multiplying both sides by m,

m'dv = F'dt        **

STUDENT QUESTION

I do not understand why m is multiplied by both sides?

INSTRUCTOR RESPONSE

The object is to get new and meaningful quantities on both sides. `dv is nothing new, (F / m) `dt is something we haven't seen before, but is recognizably the same thing as a * `dt.

m `dv and F `dt, however, are new concepts.

We call mv the momentum, so m `dv is the change in momentum.
We call F `dt the impulse.

Momentum and impulse are new and useful concepts, a significant addition to our self of 'thinking tools'.

Self-critique (if necessary):

Self-critique rating:

Question:  `qIf you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object.  What are these strategies?

Your solution: 

Confidence rating::

Given Solution: 

`a** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt.

We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **

Self-critique (if necessary):

Self-critique rating:

Question:  `qClass notes #14. 

How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?

Your solution: 

Confidence rating::

Given Solution: 

`a** a = F / m. 

vf^2 = v0^2 + 2 a `ds.  So

vf^2 = v0^2 + 2 (Fnet / m) `ds. 

Multiply by m/2 to get

1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so

Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **

STUDENT QUESTION:

Why is vf and v0 both ^2 in these equations??

INSTRUCTOR RESPONSE:

Note that the units of 2 a `ds are the same as the units for v^2. If the equation had just the first power of v it wouldn't be dimensionally consistent. It takes more than dimensional consistency to make an equation valid, but if the equation isn't dimensionally consistent the equation cannot be valid.

The reason v0 and vf are squared:

The fourth equation of uniformly accelerated motion is

vf^2 = v0^2 + 2 a `ds.

This equation was derived earlier in the course; it comes from eliminating `dt between the first two equations. The first two equations come directly from the definitions of velocity and acceleration.

Self-critique (if necessary):

Self-critique rating:

Question:  `qWhat is kinetic energy and how does it arise naturally in the process described in the previous question?

Your solution: 

Confidence rating::

Given Solution: 

`a** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. **

Self-critique (if necessary):

Self-critique rating:

Question:  `qWhat forces act on an object as it is sliding up an incline?

Your solution: 

Confidence rating::

Given Solution: 

`a**  Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp.  The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force.  Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion.

The gravitational force is conservative; all other forces in the direction of motion are nonconservative.

More rigorous reasoning:

The acceleration of the system is zero in the direction perpendicular to the incline (i.e., the object neither accelerates up and off the incline, nor into the incline).

• From this we conclude that the sum of all forces perpendicular to the incline is zero.

• In this case the only forces exerted perpendicular to the incline are the perpendicular component of the gravitational force, and the normal force.

• We conclude that the sum of these two forces must be zero, so in this case the normal force is equal and opposite to the perpendicular component of the gravitational force.

The forces parallel to the incline are the parallel component of the gravitational force and the frictional force; the latter is in the direction opposite the motion of the object along the incline.

As the object slides up the incline, the parallel component of the gravitational force and the frictional force both act down the incline.

COMMON ERROR: 

The Normal Force is in the upward direction and balances the gravitational force.

COMMENT: 

The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal.  The normal force cannot balance the gravitational force if the incline isn't horizontal.  Friction provides a component parallel to the incline and opposite to the direction of motion. **

Self-critique (if necessary):

Self-critique rating:

Question:  `qFor an object sliding a known distance along an incline how do we calculate the work done on the object by gravity?  How do we calculate the work done by the object against gravity?

Your solution: 

Confidence rating::

Given Solution: 

`a** The gravitational force is m * g directly downward, where g is the acceleration of gravity.  m * g is the weight of the object.

If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of whether the product is positive or negative. 

• If the displacement `dy is in the same direction as the weight m * g then the product is negative.

• If the displacement `dy and the weight m * g are in the same direction then the product is positive.

Alternatively it is instructive to consider the forces in the actual direction of motion along the incline.

For small inclines the magnitude of the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. 

 

The precise magnitude of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal.  This force acts down the incline. 

(You have seen that the parallel component is m g cos(270 deg - theta) or m g cos(270 deg + theta), depending on whether your incline slopes up or down as you go left to right.  These expressions follow directly from the circular definition of the trigonometric functions. 

The magnitude of cos(270 deg - theta) is the same as the magnitude of cos(270 deg + theta), and each is in turn the same as the magnitude of sin(theta).

The expression m g * sin(theta) also follows directly from the right-angle trigonometry of the situation.)

 

If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. 

• If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase.  This behavior is consistent with our experience of objects moving freely down inclines.

 

• If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative.  In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines.

The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **

NOTE ON THE EXPRESSION m g * sin(theta) 'down the incline'

Suppose the incline is at angle theta with horizontal, with the incline ascending as we move to the right.  If the x and y axes are in their traditional horizontal and vertical orientations, then the incline makes angle theta with the positive x axis, and the weight vector acts along the negative y axis.

It is more convenient to have the x axis directed along the incline, so that motion is along a single axis.  We therefore rotate the coordinate system counterclockwise through angle theta, bringing the x axis into the desired alignment.  As we do this, the  y axis also rotates through angle theta, so that the negative y axis rotates away from the weight vector.  When we have completed the rotation, the weight vector will lie in the third quadrant, making angle theta with respect to the negative y axis.  The direction of the weight vector will then be 270 deg - theta, as measured counterclockwise from the positive x axis.

The x and y components of the weight vector will then be ( m g * cos(270 deg - theta) ) and ( m g * sin(270 deg - theta) ).

It turns out that cos(270 deg- theta) = -sin(theta), and sin(270 deg - theta) = -cos(theta), so the x component of the gravitational force is -m g sin(theta); alternatively we can express this as m g sin(theta) directed down the incline.  This agrees with the given formula.

A displacement `ds up the incline (in the direction opposite the gravitational force component along the incline) implies that work `dW = -m g sin(theta) * `ds is done on the object by gravity, so that its gravitational PE increases by amount m g sin(theta) * `ds.

NOTE ON m g sin(theta) * `ds

For the same incline as discussed in the previous note, if the displacement is `ds up the incline, then the displacement vector will have magnitude `ds and will make angle theta with the horizontal.  If our x and y axes are respectively horizontal and vertical, then the displacement is represented by the vector with magnitude `ds and angle theta.  The horizontal and vertical components of this vector are respectively `ds cos(theta) and `ds sin(theta).

In particular an object which undergoes displacement `ds up the incline has a vertical, or y displacement `dy = `ds sin(theta).  This displacement is along the same line as the gravitational force m g, but in the opposite direction, so that the work done on the object by gravity is - m g * `ds sin(theta), and the change in gravitational PE is again found to be m g sin(theta) * `ds.

FOR THE PERPLEXED

If you don't understand this problem or the given solution, you should first be sure you have viewed all assigned Class Notes to date.

Then you should specifically self-critique your understanding of the following summary by copying it into your self-critique.  Alternatively you could submit these questions and your responses separately, using the Submit Work Form:

• #$*& A displacement vector of length `ds in direction theta relative to the positive x axis has components `ds * cos(theta) and `ds * sin(theta). &&

• #$*& The vertical component of the displacement vector `ds is `ds * sin(theta). &&

• #$*& If the x-y coordinate system is in the 'standard' orientation, with the x axis horizontal and the y axis vertical, then the weight vector is directed along the negative y axis.  &&

• #$*& If the coordinate system is rotated so that the x axis is directed at angle theta with above horizontal, the displacement vector whose length is `ds and whose direction makes angle theta with respect to horizontal will now point in the x direction. &&

• #$*& The weight vector lies at angle 270 degrees - theta with respect to the x axis of the rotated system. &&

• #$*& The components of the weight vector in the x and y direction are, respectively, m g cos(270 deg - theta), and m g sin(270 deg - theta). &&

• #$*& The displacement vector `ds and the weight component m g cos(270 deg - theta) are both directed along the x axis of the rotated system, the first in the positive direction and the second in the negative. &&

• #$*& When these two components are multiplied, we get the work done by the gravitational force on the mass. &&

• #$*& The vertical component `ds * sin(theta) of the displacement vector and the gravitational force m g are both in the vertical direction, one acting upward and the other downward.  So when we multiply `ds sin(theta) by (- m g) we get the work done by the gravitational force on the mass. &&

• #$*& Whichever way we choose to orient our system, we find that the work done by the gravitational force is - m g * `ds * sin(theta). &&

 #$*&

 SYNOPSIS

The work done by gravity is -m * g * `dy, where g and `dy are considered to be signed quantities (i.e., pick your positive direction and make each of these quantities + or - according to whether it is in your chosen positive direction).

The work done against gravity and the work done by gravity are equal and opposite.

For an incline at angle theta with horizontal the component of the gravitational force which acts parallel to the incline is m g sin(270 deg - theta). If this is multiplied by the displacement along the incline we get the work done by gravity.

For this same incline we have `dy = displacement along incline * sin(270 deg + theta). If we calculate m g `dy we get the same result as when we calculated mg sin(270 deg - theta).

Self-critique (if necessary):

Self-critique rating:

Question:  `q

For an object sliding a known distance along an incline how do we calculate the work done by the object against a given frictional force?  How does the work done by the net force differ from that done by gravity?

Your solution: 

Confidence rating::

Given Solution: 

`a** The work done against friction is the product of the distance moved and the frictional force.  Since the force exerted by friction is always opposed to the direction of motion, the force exerted by the system against friction is always in the direction of motion so the work done against friction is positive.

The net force on the system is sum of the gravitational component parallel to the incline and the frictional force.  The work done by the net force is therefore equal to the work done by gravity plus the work done by the frictional force (in the case of an object moving up an incline, both gravity and friction do negative work so that the object must do positive work to overcome both forces; in the case of an object moving down an incline gravity does positive work on the system while friction, as always, does negative work on the system; in the latter case depending on whether the work done by gravity on the system is greater or less than the frictional work done against the system the net work done on the system may be positive or negative) **

STUDENT QUESTION

Oops! I forgot that the frictional force is added to the force of the object. But I am a little hazy on the ‘why’ of this. Since friction works against the force of the object, wouldn’t it be subtracted from the Fnet. OR… is the friction added to the Fnet because the object has to the work of itself + the frictional component to move along the incline??

INSTRUCTOR RESPONSE Friction opposes the relative motion of two surfaces. In this case the relative motion is that of the object sliding on the surface of the incline. So the frictional force acts in the direction opposite the motion.

The gravitational force component parallel to the incline acts down the incline. So if the motion is up the incline the frictional force is in the same direction as the gravitational force. If motion is down the incline, the frictional force acts in the direction opposite the parallel gravitational component.

Self-critique (if necessary):

Self-critique rating:

Question:  `qExplain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length.

Your solution: 

Confidence rating::

Given Solution: 

`a** In terms of similar triangles:

The reason the approximation only works for small displacements is because the sides used on one triangle are not the same as the sides used on the other.  From the triangle we see that the restoring force and the weight are at right angles, while the length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle and hence are not at right angles.

For small angles the two long sides of the triangle are approximately equal so the discrepancy doesn't make much difference.  For larger angles where the two long sides are significantly different in length, the approximation no longer works so well.

In terms of components of the vectors:

The tension force is in the direction of the string.

The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement (just draw the picture).

The vertical component of the tension force must be equal to the weight of the pendulum, since the pendulum is in equilibrium.

If the displacement is small compared to the length the vertical component of the tension force will be very nearly equal to the tension force.  So the previous statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement' can be replaced by the statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the weight of the pendulum as the length of the pendulum to the horizontal displacement. **

This situation is illustrated in the figures below.  Note that this is also explained in video embedded in the DVD version of the Class Notes.

The first figure depicts a simple pendulum consisting of a symmetric mass suspended from a fixed point by a light string.  The pendulum is not in its vertical position, but is displaced a bit to the right of that position.

The string exerts a force that prevents the pendulum from accelerating vertically downward.  This force results when the string is stretched slightly in response to the weight of the pendulum, and is called a tension force.  It acts along the line of the string, pulling the pendulum up and toward the left.

The weight of the pendulum is the force exerted on it by gravity.  The weight acts in the vertical downward direction.

The figure below depicts the tension and the weight.

The next figure depicts only the tension force.

In the following picture we superimpose a triangle on the preceding picture.  The hypotenuse of the triangle coincides with the tension vector.  The legs of the triangle are in the horizontal and vertical directions.

The triangle is shown by itself below.  The length of the hypotenuse represents magnitude of the tension force, the lengths of the vertical and horizontal legs represent the vertical and horizontal components of the tension vector.

The vertical leg and hypotenuse are of very nearly the same length.  Therefore the vertical component of the tension is very nearly equal in magnitude to the tension.

As the pendulum swings back toward equilibrium it moves almost entirely in the horizontal direction, and therefore has practically no vertical acceleration.  The net vertical force is therefore practically zero.  We conclude that the vertical component of the tension is very nearly equal and opposite the weight of the pendulum.

In terms of the notation of the next figure, in which the sides of the triangle are labeled in terms of the tension and its components, we see that

T_y is equal to the weight of the pendulum

T is very nearly the same as T_y

T_x is the horizontal component of the tension. 

• In this figure T_x appears to be about 1/10 as great as T_y.

• So we say that the ratio T_x / T_y is roughly 1/10, or .1.

Since T_y is equal to the weight of the pendulum, T_x / T_y is the ratio of the x component of the tension to its weight:

• T_x / T_y = T_x / weight

In the next figure we superimpose a similar triangle on the original sketch of the pendulum.  The length of the triangle is equal to the length of the pendulum, and the horizontal leg is the displacement of the pendulum from its equilibrium position.

Sketching the triangle by itself and labeling its hypotenuse L (for the length of the pendulum) and its horizontal leg x we have the figure below:

The vertical leg of this triangle is very nearly the same length as L.  So the ratio of horizontal to vertical legs is very close to x / L.

This triangle is geometrically similar to the triangle we used previously to represent the components of the tension.  The geometric similarity implies that the ratio of horizontal to vertical leg must be the same for both.  Writing this condition in symbols we have

T_x / T_y = x / L

Since T_y is nearly the same as the weight we have

T_x / weight = x / L.

That is, the force restoring the pendulum to equilibrium is in the same proportion to the weight of the pendulum as the displacement from equilibrium to its length.

Note that this topic was covered in Class Notes #14, which are part of the preceding assignment.

Self-critique (if necessary):

Self-critique rating:

Question (openstax):  7.5

`prin, `gen:  Calculate the work done by an 75-kg man who pushes a crate 5 m up along a ramp that makes an angle of 15º with the horizontal.

(See Figure 7.35.) He exerts a force of 400 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he

does on the crate and on his body to get up the ramp.

(Notes:  Assume that the man's effort starts at this point; don't worry about how the crate got to the pictured position.  If you know the trigonometry, use it, but otherwise understand that the force exerted by the man parallel to his direction of motion up the 15 degree incline is about 25% of his weight.)

Your solution: 

Confidence rating:

Given Solution: 

The man must move the crate up the ramp, and must also move himself up the ramp.

To move the crate up the ramp he must exert the 400 N force through a displacement of 5 meters in the direction of the force, so he must do work

`dW_crate_ON = 400 N * 5 m = 2000 Joules.

(note that the subscripts 'crate' and 'ON' indicate that the work calculated here is done on the crate).

To move himself up the incline he must exert the necessary force in the direction of the incline.  That force is equal to about 25% of his weight.  His weight is 75 kg * 9.8 m/s^2 = 730 Newtons, approx., so the required force in the direction up the incline is about 25% of this, or roughly 180 Newtons.  So the work required to move himself up the incline is

`dW_man = 180 N * 5 m = 900 Joules.

The total work required is therefore

`dW = 2000 J + 900 J = 2900 J.

Self-critique (if necessary):

Self-critique rating:

Question:  `qprin and gen phy:  6.4:  work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.

Your solution: 

Confidence rating::

Given Solution: 

`aThe net force on the crate must be zero, since it is not accelerating.  The gravitational force on the crate is 160 kg * 9.8 m/s^2 = 1570 N, approx.  The only other vertical force is the normal force, which must therefore be equal and opposite to the gravitational force.

As it slides across the floor the crate experiences a frictional force, opposite its direction of motion, which is equal to mu * normal force, or .50 * 1570 N = 780 N, approx..  The only other horizontal force is exerted by the movers, and since the net force on the crate is zero the movers must be exerting a force of 780 N in the direction of motion.

The work the movers do in 10.3 m is therefore

work = Fnet * `ds = 780 N * 10.3 m = 8000 N m = 8000 J, approx..

Self-critique (if necessary):

Self-critique rating:

Question (openstax): `gen, `prin 7.4 

The ski patrol lowers a rescue sled and victim, having a total mass of 80 kg, down a 60º slope at constant speed. 

The frictional force is 7% of the normal force between the skier and the slope.  The weight of the skier has components in the direction of the slope, and perpendicular to the direction of the slope, the former being about 87% of her weight and the latter 50%.  The net force perpendicular to the slope is zero.

Your solution: 

Confidence rating:

Given Solution: 

Self-critique (if necessary):

Self-critique rating:

Question:  `qgen phy prob 6.9:  force and work accelerating helicopter mass M at .10 g upward thru dist h.

Your solution: 

Confidence rating::

Given Solution: 

`aTo accelerate the helicopter at .10 g it must experience net force Fnet = mass * acceleration = M * .10 g = .10 M g.

The forces acting on the helicopter are its upward thrust T and the downward pull - M g of gravity, so the net force is T - M g.  Thus we have

T - M g = .10 M g, and the upward thrust is

T = .10 M g + M g = 1.10 M g.

To exert this force through an upward displacement h would therefore require

work = force * displacement = 1.10 M g * h = 1.10 M g h.

STUDENT COMMENT AND INSTRUCTOR RESPONSE:  I didn't think of that. I still don't fully understand it.

INSTRUCTOR RESPONSE: 

F_net = m a = m * .10 g = .10 m g.

F_net = upward thrust + gravitational force = T - m g.

Thus T - m g = .10 m g.

STUDENT QUESTION

I totally spaced on the Thrust aspect of this problem. What indicators are there in a problem that tell me to incorporate Thrust, when it doesn’t state it within the problem. Do I have to use it in all problems when things are lifted off the ground??

INSTRUCTOR RESPONSE

If it's accelerating upward at .10 g, then the net force is .10 M g. It doesn't matter what 'it' is; could be a helicopter, an elevator, a fish on a line.

Since gravity is pulling it downward something, whatever we wish to call it, is pulling it upward, with a force we might as well represent by the letter T (any other letter would do as well; we could revert to the old standby and call the force x).

• It doesn't hurt to have a word for this force, but we don't really need one; all we really need is to know it's there.

• It's not necessary to call this force 'thrust' (applicable to the helicopter) or 'tension' (applicable to the elevator or the fish, which are being pulled upward by the tension in a cable or fishing line).

• But whatever we want to call it (if anything), or whatever symbol we want to use for it, we know it has to be there because if it wasn't the object would be accelerating downward at 9.8 m/s^2.

Using T for the unknown force, the net force on the object is T - M g.

This gives us two expressions for the net force. We set these two expressions equal and get the equation

T - M g = .10 M g

and solve to get T = 1.10 M g

Self-critique (if necessary):

Self-critique rating:

Question:  `q****   Univ:  6.58 (6.50 10th edition).  chin-up .40 m, 70 J/kg of muscle mass, % of body mass in pullup muscles of can do just 1.  Same info for son whose arms half as long.

Your solution: 

Confidence rating::

Given Solution: 

`a** For each kg of mass the weight is 1 kg * 9.8 m/s^2 = 9.8 N.

Work done to lift each kg of mass .4 m would then be 9.8 N * .4 m = 3.92 J.

The chin-up muscles generate 3.92 J per kg, which is 3.92 / 70 of the work one kg of muscle mass would produce.  So the proportion of body mass in the pullup muscles is 3.92 / 70 = .056, or 5.6%.

For the son each kg is lifted only half as far so the son only has to do half the work per kg, or 1.96 J per kg.

For the son the proportion of muscle mass is therefore only 1.96 / 70 = 2.8%.

The son's advantage is the fact that he is lifting his weight half as high, requiring only half the work per kg.   **

Self-critique (if necessary):

Self-critique rating:

Question: (NOTE:  THIS PROBLEM HAS BEEN OMITTED FROM THE TEXT AND IS NOT PRESENTLY ASSIGNED)  `q Univ.  6.72 (6.62 10th edition).  A net force of 5.00 N m^2 / x^2 is directed at 31 degrees relative to the x axis.  ; obj moves along x axis, mass .250 kg, vel at x=1.00 m is 4.00 m/s so what is velocity at x = 1.50 m?

We first find the component in the x direction which would be

F_x = 5.00 N m^2 / x^2 * cos(31 deg)
= 4.25 N m^2 / x^2, approx..

We would then integrate this force from x = 1.00 m to x = 1.50 m. An antiderivative would be

-4.25 N m^2 / x

The change in this antiderivative over the interval of integration would be

-4.25 N m^2 ( 1 / (1.50 m) - 1 / (1.00 m) ) = -4.25 N m^2 * (-1/3 m^-1) = 1.42 N m = 1.42 J.
 

Initial KE is 1/2 * .250 kg * (4.00 m/s)^2 = 2 J
Final KE is 1/2 * .250 kg * (4.00 m/s)^2 + 1.4 J = 3.4 J, approx. so final vel is
vf = sqrt( 2 KEf / m) = sqrt( 2 * 3.4 J / (.250 kg) ) = sqrt( 27 m^2 / s^2) = 5.2 m/s, approx..

SOLUTION FOR A DIFFERENT FORCE FUNCTION

Here is a solution for net force function 5 N / m^2 * x^2.

the component of the force in the direction of motion is 5.00 N / m^2 * x^2 * cos(31 deg) = 5.00 N / m^2 * x^2 * .86 = 4.3 N/m^2 * x^2.

Integrating this with respect to x from x = 1.00 m to x = 1.50 m we get something around 3 Joules (antiderivative is 4.3 N / m^2 * x^3 / 3 = 1.43 N/m^2 * x^3; the change in the antiderivative is about 1.43 N/m^2 [ ( 1.50 m)^3 - (1.00 m)^3 ] = about 3 N * m = 3 J).

Initial KE is 1/2 * .250 kg * (4.00 m/s)^2 = 2 J
Final KE is 1/2 * .250 kg * (4.00 m/s)^2 + 3 J = 5 J, approx. so final vel is
vf = sqrt( 2 KEf / m) = sqrt( 2 * 5 J / (.250 kg) ) = sqrt( 40 m^2 / s^2) = 6.4 m/s, approx..

Self-critique (if necessary):

Self-critique rating:

Question:  (University Physics students only)  What is the work done by force F(x) = - k / x^2 between x = x1 and x = x2.

Your solution: 

Confidence rating::

`a** Force is variable so you have to integrate force with respect to position.

Position is measured along the x axis, so you integrate F(x) = - k / x^2 with respect to x from x1 to x2.

An antiderivative of - k / x^2 is k / x so the integral is k / x2 - k / x1.

If x2 > x1, then k / x2 < k / x1 and the work is negative.

Also, if x2 > x1, then motion is in the positive x direction while F = - k / x^2 is in the negative direction.  Force and displacement in opposite directions imply negative work by the force.

For slow motion acceleration is negligible so the net force is practically zero.

Thus the force exerted by your hand is equal and opposite to the force F = - k / x^2.

The work you do is opposite to the work done by the force so will be - (k / x2 - k / x1) = k/x1 - k/x2, which is positive if x2 > x1.

This is consistent with the fact that the force you exert is in the opposite direction to the force, therefore in the positive direction, as is the displacement.

Note that the work done by the force is equal and opposite to the work done against the force. 

**

Self-critique (if necessary):

Self-critique rating:

Calculus-related comments and observations (applicable mainly to University Physics; may be of interest to students in other physics courses who know calculus):

Good observation by student:

the derivative of change in KE is mvf-mv0. so i suppose this is the change in momentum...?

Instructor response:

Good observation.

It's important to be very specific about derivatives (e.g., to specify the derivative of what with respect to what):

Strictly speaking, for a given interval 1/2 m vf^2 - 1/2 m v0^2 is a fixed quantity, so its derivative is zero.  However KE typically changes from one instant to another, and can be regarded as a continuous function of velocity.

Specifically KE = 1 / 2 m v^2, where v can vary, so the derivative of this expression with respect to v is m v:

• d(KE)/dv = m v = momentum 

In words, the derivative of KE with respect to velocity is momentum. Thus

• momentum is the rate of change of KE with respect to v.

We can go even further:

Of course velocity can be regarded as a function of clock time.  The chain rule tells us that

• d(KE) / dt = d(KE) / dv * dv / dt.

As we have seen, d(KE)/dv = m v = momentum; and we know that dv/dt = a, the acceleration.  So using a(t) to explicitly express the acceleration as a function of clock time, we have

• d(KE) / dt = (m v) * a(t) = momentum * acceleration.

The derivative of the KE with respect to clock time is the product of momentum and acceleration.

Integrating the force function for a pendulum or a spring:

The force restoring a pendulum to equilibrium (provided displacement from equilibrium is small compared to length), or to restore a spring to its equilibrium position, is of the form

• F = - k x. 

This force is conservative in the ideal case.

The work done by this force is found by integrating the force with respect to position x.

• Integrating with respect to x, the force constant k is a constant.

• The force function is F(x) = - k x; integrating this function with respect to x we get - k x^2 / 2 + c, where c is an integration constant.

• If x = 0 is taken as the 0-energy point (appropriate to the equilibrium point) then -k * 0^2 / 2 + c = 0, so c = 0 and the work done by the system against the conservative force is - k x^2 / 2.

• The potential energy being the negative of the work done against the conservative force is

PE = k x^2 / 2.