class 091102

Class 091102

Things you need to know for the test:

Here is an overview of things you really need to understand for the upcoming test:

Definitions of rate, velocity, acceleration.
Equations of uniformly accelerated motion.
Newton's three laws.
Definitions of work/energy, KE, impulse, momentum.
Work-KE theorem, definition of PE, work-energy theorem
Impulse-momentum theorem, conservation of momentum.
Vector components from magnitude and angle; magnitude and angle from vector components.

and, of course

Ability to apply the above to various situations.

Using your notes from class, information given in class notes, information given in the text, tt is suggested that you carefully write out what you know, and are supposed to know, about each of the first seven topics above.

Short-answer questions

`q001. What are the magnitude and the standard angle of a vector whose x and y components are respectively -7 m/s and +5 m/s?

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`q002. If you add a vector with magnitude 3 N and angle 70 degrees to a vector with magnitude 5 N and angle 210 deg, what are the components of the resultant?

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What are the magnitude and angle of the resultant? (note: the resultant of a set of vectors is their sum, i.e., what you get when you add them).

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`q003. What are the x and y components of the initial projectile velocity of a pendulum mass if you let go of the string when the mass is traveling at 80 cm/s at an angle of 15 degrees above horizontal (i.e., what are the x and y components of a velocity vector with magnitude 80 cm/s and angle 15 degrees as measured from a horizontal x axis)?

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Additional problems relevant to preparation for the test:

`q004. An ideal projectile is initially at a known distance h above the floor, with vertical velocity v0_y and horizontal velocity v0_x. What quantities are known for its vertical motion?

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Can these quantities be used to figure out the time interval `dt between the initial instant and contact with the floor? If so, how?

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Having found `dt, is it possible to find the distance the projectile moves in the horizontal direction during this interval? If so, how?

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`q005. A ball of mass .4 kg is tossed vertically upward at 12 m/s and a point which is 3 meters above the ground. It rises a certain distance then falls back down, where it is caught at a point 1 meter above the ground.

What is its KE at the initial instant?

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What is `dPE between the initial and final instant?

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What therefore is the KE at the final instant, and how fast is it therefore traveling? Assume that air resistance is negligible. Solve this problem using energy considerations only, without analyzing the ball's uniformly accelerated motion.

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`q006. At the instant it stops at the top of its path, by how much has the gravitational PE of the ball in the preceding exercise changed?

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What therefore is its height above the ground at that instant? Solve this problem using energy considerations only, without analyzing the ball's uniformly accelerated motion.

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`q007. Sketch the forces acting on a steel ball as it rolls down a steel ramp which is sloped in such a way that its velocity remains constant. What is the net force on the ball as it rolls down this ramp, and how does your sketch depict this fact?

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Sketch the forces acting on the same ball as it rolls up this ramp. What is the net force on the ball as it rolls up the ramp, and how does your sketch depict this fact?

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Optional derivation of equations of uniformly accelerated motion using calculus

For reference by students with a calculus background, recommended for students who have more physics in their future.

The equations of uniformly accelerated motion are obtained using the concept of the derivative as follows:

v = dx/dt (velocity is the derivative with respect to clock time of position x, the instantaneous rate of change of y with respect to x)

a = dv/dt (acceleration is the instantaneous rate of change of velocity with respect to clock time)

If a = constant, then dv/dt = a = constant so v = v0 + at (check it out that the t derivative of v0 + a t is a).

If v = v0 + a t then dx/dt = v0 + a t so x = x0 + v0 t + 1/2 a t^2 (check it out that the t derivative of x0 + v0 t + 1/2 a t^2 is v0 + a t)

From the calculus perspective, then, the two most fundamental equations of uniformly accelerated motion are

v = v0 + a t and

x = x0 + v0 t + 1/2 a t^2.

If we eliminate t and rearrange the result we get

v^2 = v0^2 + 2 a ( x - x0) and

x = x0 + (v + v0) / 2 * t.

Some general comments in response to recent work received:

For a ball of weight .59 N rolling 50 cm along an incline: The weight doesn't act in the direction of the .5 m displacement. Only the component of the weight in the direction of the displacement is multiplied by the displacement.
Friction acts in opposite directions as a ball rolls in opposite directions, up and down a sloped ramp. This is the reason for the difference in the magnitudes of the observed accelerations.
If the momentum change of one object in a collision is positive, the momentum change of the other is negative.
The direction of the vector is specified by its angle theta as measured in the counterlockwise direction from the positive x axis.
If the graph of acceleration vs. mass, for a fixed net force, is linear and decreasing, then there would be a finite mass at which acceleration reached zero. That would imply F_net = 0. So the graph can't be linear.
If F_net = m * a is constant, then a graph of a vs. m is curved, asymptotic to both vertical and horizontal axes
When a rubber band chain is stretched from length 50 cm, where it exerts a force of 4 N, to a length of 80 cm, where it exerts a force of 12 N:
4 N is not the average force for the .5 m stretch, nor is 12 N the average force for the .8 m stretch. These are the forces when the rubber band reaches these lengths, not the average forces exerted in the process of reaching the given lengths. The chain exerts an average force of approximately (4 N + 12 N) / 2 = 8 N, through a displacement of .8 m - .5 m = .3 m, which requires work 8 N * .3 m = 2.4 Joules.
If one coin accelerates the system at 5 cm/s^2, then it would take a force equal to the force of gravity on 200 coins to accelerate the system at 1000 cm/s^2. This is because acceleration is proportional to net force, and 1000 cm/s^2 / (5 cm/s^2) = 200.
To add two vectors:

add all the x components to get the x component of the resultant
add all the y components to get the y component of the resultant
find the magnitude and angle of the resultant

sketch forces on const-vel ramp and on level ramp, understand how diff in accel in opp directions is related to accel of rolling friction (which is half the former); good problem: show why the difference in the accelerations due to rolling friction is harder to measure on a steeper ramp, but unreliable on a level real-world ramp (due to increased nonuniformity of acceleration due to irregularities); wonder if the aluminum ramp would be better (more precisely machined), also if coeff of friction between steel and aluminum differs by a measurable amount

`q008: Argue for the truth or falsity of the following statement: 'When rolling down a constant-velocity ramp, the rolling friction and the 'parallel component' of the gravitational force are equal and opposite'

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Inclines, etc..

In conjunction with class discussion and the force diagrams as sketched on the board:

A weight on an incline is generally analyzed in terms of a coordinate system in which the x axis is directed along the incline, with the y axis perpendicular to the incline. This has the following advantages:

If the object moves along the incline, the y coordinate of its position remains 0, so that y velocity is always 0, as it y acceleration. So the net force in the y direction is always zero. All the force components in the y direction therefore add up to zero. One of these forces is the normal force, which can therefore be found if the y components of all other forces are known.
The net force is in the x direction, and is the sum of the x components of all the forces acting on the object.
Frictional forces act in the x direction, and are limited by mu * F_normal (coefficient of friction multiplied by normal force).

For a ball rolling along an incline at angle alpha with horizontal, subject to gravitational and normal forces and the the force of rolling friction:

the weight vector makes angle alpha with the downward y axis so its angle is theta = 270 deg + alpha or theta = 270 deg - alpha, depending on whether the incline is depicted as sloping down and to the right, or down and to the left
the ball's weight m g is represented by the components m g_x and m g_y, where m g_x = m g cos(theta) and m g_y = m g sin(theta)
the normal force exerted by the incline on the ball is in the positive y direction, and is equal and opposite to m g_y; this makes the net force in the y direction zero
the frictional force is mu_rolling * | F_normal |, in the direction opposite the ball's motion
If we have sketched the figure sloping down and to the right, so that the direction down the ramp is the positive x direction, then we can say the following:

theta = 270 deg + alpha and wt_x = m g_x = m g cos(theta) (this is the 'parallel component of the weight).

If the ball is rolling down the incline then its velocity is positive, friction is up the incline and hence negative and the frictional force, whose magnitude is mu_rolling * | F_normal |, is negative. The net force on the ball is therefore

F_net = m g_x - mu_rolling * | F_normal |.

In terms of the angle alpha of the incline with horizontal, mu_rolling and m we have

F_net = m g cos(270 deg + alpha) - mu_rolling * | m g sin(270 deg + alpha) |

The point here is not to memorize this equation, but to recognize in each expression the way it emerged from our sketch.

Summarizing:

270 deg + alpha is the angle of the weight vector measured counterclockwise from the positive x axis

m g cos(270 deg + alpha) and mg sin(270 deg + alpha) are the x and y components of the gravitational force on the mass m

the normal force is in this case equal and opposite to the y component of the gravitational force

and mu_rolling * | m g sin(270 deg + alpha) | is the cofficient of rolling friction multiplied by the normal force

If the ball is rolling up the incline, then the parallel component of the gravitational force remains the same but the frictional force will be in the opposite direction and our net force becomes

F_net = m g cos(270 deg + alpha) + mu_rolling * | m g sin(270 deg + alpha) |

Had we sketched the figure sloping down and to the left, the direction down the ramp would have been in the negative x direction. So theta would be 270 deg - alpha (which would make cos(theta) negative).

A ball rolling down the incline would be moving in the negative x direction so friction would be in the positive x direction and we would have

F_net = m g cos(270 deg - alpha) + mu_rolling * | m g sin(270 deg + alpha) |

A ball rolling up the incline would be moving in the positive x direction so friction would be in the negative x direction and we would have

F_net = m g cos(270 deg - alpha) - mu_rolling * | m g sin(270 deg + alpha) |

Once more, all these equations for F_net are confusing. The expression depends on how you sketch the figure and whether the ball is moving up or down the incline. If you make a good sketch, and remember that the frictional force is directed opposite to the motion of the object, the reasoning process should lead you to the correct expression. It would not be a bad idea at this point to make four sketches, one representing each of these situations, and see if you get the correct expressions for F_net.

For a mass m1 on a level surface (think of a table), attached by a string to a mass m2 suspended over a pulley:

m1 rests on a level surface, which supports it. The normal force on this mass is equal and opposite to the gravitational force for |F_normal| = m g and the two together make no contribution to the net force on the two-mass system.
m1 is subjected to a frictional force f_frict < = mu * | F_normal |.
m2 is subjected to the downward force m2 g of gravity

We will regard both masses as one system. Its total mass is m1 + m2.

We will take the positive direction as the direction in which the hanging mass descends.

The string connecting the masses exerts a tension force on each mass, one tension force acting in the positive direction and the other in the negative. Being equal and opposite, these forces contribute nothing to the net force on the system. They can be regarded as internal forces, forces which act within but not on the system as a whole.

The net force acting on the system is m2 g, in our chosen positive direction, and f_frict. The direction of the frictional force depends on whether the system is moving and if so, in what direction.

The behavior of this system depends on whether it is initially stationary, and if it is initially moving, in which direction.

If it is initially stationary, then if f_frict >= m2 g the system will remain stationary. Note that in this case the coefficient of friction is the coefficient of static friction, so f_friction <= mu_s * m1 g. Thus our condition for remaining stationary would be

m2 g <= mu_s * m1 g, which could be simplified to

m2 <= mu_s * m1

If the system is initially moving in the positive direction, then the frictional force is in the negative direction, and the coefficient of friction is the coefficient of kinetic friction mu_k. Thus our net force is

F_net = m2 g - mu_k * m1.

If mu_k * m1 is greater than m2 g, the net force will be negative and the moving system will slow down. If mu_k * m1 = m2 g, then the net force will be zero and the system will be in equilibrium, moving at a constant velocity (at least until m2 reaches the edge of the table).

If the system is initially moving in the negative direction, then the frictional force is in the positive direction and

F_net = m2 g + mu_k * m1.

The net force and therefore the acceleration are positive, and the system, which is moving in the negative direction, will slow down.

`q009. If a mass of 5 kg on a level surface is attached by a horizontal string to a mass of 2 kg suspended over a pulley, then

If there is no friction what is the acceleration of the system?

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The net force on the system is the force of gravity on the 2 kg mass, which is 2 kg * 9.8 m/s^2 = 19.6 N.

The mass of the system is 5 kg + 2 kg = 7 kg.

The acceleration of the system is therefore

a = F_net / m = 19.6 N / (7 kg) = 2.8 m/s^2.

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If the coefficient of static friction is .3 and the coefficient of kinetic friction is .2, what would be the acceleration of the system if released from rest?

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The normal force supporting the 5 kg mass is equal and opposite to its weight, so

| F_normal | = 5 kg * 9.8 m/s^2 = 49 N.

The force of static friction has magnitude

| f_friction_stat | < = mu_stat * | F_normal | = .3 * 49 N = 14.7 N.

This is less than the 19.6 N force of gravity on the suspended mass, so static friction will exert the full 14.7 N force to to prevent motion, but it won't be enough.

The object will then experience kinetic friction of magnitude

| f_friction_kin | = .2 * 49 N = 9.8 N

in the direction opposite motion.

Taking as positive the direction in which the suspended mass descends, we have

F_net = 19.6 N - 9.8 N = 9.8 N so that

a = F_net / m = 9.8 N / (7 kg) = 1.4 m/s^2.

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If the coefficient of static friction is .3 and the coefficient of kinetic friction is .2, what would be the acceleration of the system if at a certain instant was moving at 2 m/s in the direction which raises the second mass? In this case, would the system eventually return back to its initial position (assume the tabletop and string are long enough that nothing hits the pulley or falls off the table), and if so how long would this take and how fast would the system be moving at the instant of return? How much energy would have been dissipated to friction?

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Homework:

Your label for this assignment:

ic_class_091102

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form. Answer the questions posed above.

You should know everything in the first six problems of Set 5 in the Introductory Problem Set, which will give you a good, and not difficult, introduction to working with vectors. A link that should get you there is at http://vhmthphy.vhcc.edu/ph1introsets/default.htm .

You should also read Chapters 4 and 6 in your text, which can serve as a good coherent reference. You should make note of some of the differences in notation (e.g., my use of `dW_NC_ON where the book uses just W_NC).

Practice Tests

Principles of Physics (Phy 121) Test_1

Problem Number 1

Two objects collide and remain stuck together after collision.

One object has mass 9 kg and is moving in the positive direction at 20 m/s and the other has mass 7 kg and and moves at 16 m/s in the negative direction.

What is their total momentum, and what will be the velocity of this system immediately after collision?

Total momentum = 9 kg * 20 m/s + 7 kg * (-16 m/s)

Total mass = 9 kg + 7 kg = 16 kg

Momentum after collision = momentum before collision

both masses move at the same velocity after collision so

momentum after collision = total mass * common velocity.

Therefore common velocity = momentum after collision / total mass.

Generalizing:

In symbols, with v1 and v2 the before-collision velocities and v the common after-collision velocity:

p_before = m1 v1 + m2 v2

p_after = (m1 + m2) v

p_after = p_before so

(m1 + m2) v = m1 v1 + m2 v2

Problem Number 2

You walk along level ground in a straight line from some initial point to some terminal point.

If I walk 2.37 miles to the East, then I will be directly South of your terminal point. Your terminal point will lie 3.04 miles to the North.

What angle does your path make with East?
How far did you walk?

If I turn north and walk 3.04 miles to your terminal point, my path follows the legs of a right triangle, 2.37 miles along one leg (to the east) and and 3.04 miles along the other (to the north).

You walk along the hypotenuse.

So you walk distance

sqrt( (2.37 miles)^2 + (3.04 miles)^2 )

at an angle of

theta = arcTan(3.04 miles / (2.37 miles) )

as measured counterclockwise from the easterly direction.

You would of course complete these calculations.

Problem Number 3

What are x and y the components of the velocity vector obtained when we add the two following velocity vectors:

vector A, with x and y components 7.1 m / s and 3.6 m / s, and
vector B whose x and y components are 6.4 m / s and -6.2 m / s?

What are the magnitude and angle of the resultant vector?

The x components add up to R_x = 7.1 m/s + 6.4 m/s = 13.5 m/s.

The y components add up to R_y = 3.6 m/s + (-6.2 m/s) = -2.6 m/s.

The magnitude of the resultant is sqrt( R_x^2 + R_y^2) and its angle is arcTan(R_y / R_x).

R_x is in this case positive so we don't add 180 deg.

However the arcTan turns out to be negative. To get the angle with the positive x axis we add 360 deg.

Problem Number 4

An object with mass 15 kilograms, initially at rest, is acted upon by a net force of 4650 Newtons.

If the force acts for 9 seconds, what will be KE increase of the object, based on its velocity change?
During the 9 seconds, based on the distance it travels, how much work is done by the net force on the object?

The acceleration of the mass will be a = Fnet / m = 4650 N / (15 kg) = 310 m/s^2.

In 9 sec velocity will therefore change by

`dv = a `dt = 310 m/s^2 * 9 s = 2800 m/s, approx..

It starts from rest so its initial velocity is 0, its final velocity about 2800 m/s.

Its KE will therefore increase from

KE_0 = 1/2 * 15 kg * (0 m/s)^2 = 0 J

KE_f = 1/2 * 15 kg * (2800 m/s)^2 = 60 000 000 J, approx..

So the KE increase will be about

`dKE = KE_f - KE_0 = 60 000 000 J - 0 J = 60 000 000 J.

So its average velocity is about

vAve = (0 m/s + 2800 m/s) / 2= 1400 m/s.

In 9 seconds its displacement will therefore be about

`ds = vAve * `dt = 1400 m/s * 9 s = 13600 m.

A force net of 4650 N acting through a displacement of 13 600 m does work

`dW_net = 4650 N * 13 600 m = 60 000 000 Joules, approx.

in agreement with `dKE.

This verifies for this case that

`dW_net = `dKE.

Problem Number 5

What are the magnitude and angle of a vector whose x and y components are respectively 5.8 and 7.5?

The magnitude is sqrt(5.8^2 + 7.5^2). If you work this out it comes to roughly 9.5.

The angle with the positive x axis is

arcTan(7.5 / 5.8),

which is around 50 degrees.

Problem Number 6

How far must an object be pushed by a force of 9 Newtons, always exerted parallel to the direction of motion, in order to gain 16 Joules of energy, if the energy gained is equal to the work done on the object?

Since `dW = F `ds, we have

`ds = `dW / F = 16 Joules / (9 Newtons) = 16 kg m^2 / s^2 / (9 kg m/s) = 1.8 meters, roughly.

Problem Number 7

Over an interval of .05 seconds, the velocity of an object of constant mass 3 Kg is observed to change from 7 m/s to 1.9 m/s.

Find the average force on the object using the Impulse-Momentum Theorem.
Verify your results using your knowledge of uniformly accelerated motion.

Using impulse-momentum:

F_net_ave `dt = `d(m v).

m v changes from 3 kg * 7 m/s = 21 kg m/s to 3 kg * 1.9 m/s = 5.7 kg m/s, so

`d(mv) = 5.7 kg m/s - 21 kg m/s = -15.3 kg m/s.

The average net force is therefore

F_net_ave = `d(m v) / `dt = -15.3 kg m/s / (.05 s) = 306 kg m/s^2, or 306 Newtons.

Using uniformly accelerated motion:

The average acceleration is

a_ave = `dv / `dt = (1.9 m/s - 7 m/s) / (.05 s) = -102 N, so by Newton's Second Law

F_net_ave = m * a_ave = 3 kg * (-102 N) = -306 N.

Problem Number 8

An object of mass 10 kg experiences a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for 2 seconds.

If the average force over this time is 9800 Newtons,

Use Newton's Second Law and your knowledge of uniformly accelerated motion to find the change in the object's velocity.
Use the Impulse-Momentum Theorem to obtain the same result.

This is very similar in its solution to the preceding problem.

Problem Number 9

Assuming that 12096 Joules of kinetic energy are dissipated in the process, how much kinetic energy increase should be expected as a result of a 90 Newton force being exerted over a distance of 160 meters?

90 N acting through a displacement of 160 meters does work

`dW = 90 N * 160 m = 14 400 Joules

If 12 100 J are dissipated then the kinetic energy change is

`dKE = work done - dissipated energy = 14 400 J - 12 100 J = 2300 J.

This problem isn't very specific about the nature of the 90 N force; it implicitly makes the simplest assumptions, e.g., that none of the 90 N force is dissipative, and that there is no change in potential energy.

General College Physics (Phy 201) Test 1

Problem Number 1

A block of mass 6 kg on an incline angled at 24 degrees above the horizontal experiences a frictional force resisting its tendency to slide along the ramp. The upper limit on this frictional force is .13 of the normal force between the block and the ramp. What force must be exerted parallel to the ramp so that the block will slide at a uniform velocity up the ramp?

Sketch your diagram and impose your x-y axes appropriately.

What are the components of the weight parallel and perpendicular to the incline?

What therefore is the normal force?

What therefore is the frictional force?

What is the net force (hint: uniform velocity implies zero acceleration)?

What is the sum of the gravitational and frictional forces?

What additional force is therefore required to get the net force you specified earlier?

Problem Number 2

The velocity of an object increases at a uniform rate from 9 m/s to 27 m/s in 6 seconds.

How far does it travel and what is its acceleration?

Sketch the corresponding velocity vs. clock time graph.

Explain the meaning of the slope of the graph and its area.

What is the average velocity of the object?

How far does it therefore travel in the given time interval?

What is the change in its velocity? What therefore is its acceleration?

The v vs. t graph forms a trapezoid. What are its altitudes, its width, its rise, its run, its slope, the dimensions of the equal-area rectangle, the corresponding area, and what does each of these quantities represent?

Problem Number 3

A cart of mass 1.2 kg coasts 55 cm down an incline at 4 degrees with horizontal. Assume that the frictional force is .049 times the normal force, and that other nongravitational forces parallel to the incline are negligible.

What is the component of the cart's weight parallel to the incline?
How much work does this force do as the cart rolls down the incline?
How much work does the net force do as the cart rolls down the incline?
Using the definition of kinetic energy determine the velocity of the cart after coasting the 55 cm, assuming its initial velocity to be zero.

Sketch your diagram and impose your x-y axes appropriately.

What are the components of the weight parallel and perpendicular to the incline?

What therefore is the normal force?

What therefore is the frictional force?

Through what displacement does the parallel component of the weight act, and how much work does it do?

What is the net force, through what displacement parallel to the force does it act, and how much work does it do?

What therefore is the change in KE?

What therefore is the change in velocity?

Problem Number 4

A system consists of a cart pulled along a constant-velocity ramp by the force of gravity on a single paper clip, whose mass is much less than that of the cart, attached by a thread over a pulley with negligible friction. If the system accelerates at 4.9 cm/s², and if F = m a describes the relationship among net force F, mass m and acceleration a, give the acceleration of each of the following:

The same system but with 6 paper clips instead of one.

The same system but with a single paper clip and a cart of twice the mass.

The same system but with a single paper clip with a cart of half the mass.

The same system but with 9 paper clips and a cart of 19 times the mass.

What would be the acceleration of the same system but with a number of paper clips whose mass equals that of cart?

How would the slope of a graph of total paper clip weight vs. acceleration for the original system (for a small number of paper clips) compare with a slope of a similar graph for a system with half the cart mass, and how would it compare for the slope of a system with 9 times the cart mass?

Will the additional paper clips significantly affect the mass of the system, and if so how? Will they significantly affect the net force accelerating the system, and if so how? Since a = F_net / m, what is your conclusion about the effect on the acceleration?

Will doubling the mass of the cart significantly affect the mass of the system, and if so how? Will doubling the mass of the cart significantly affect the net force accelerating the system, and if so how? Since a = F_net / m, what is your conclusion about the effect on the acceleration?

Will halving the mass of the cart significantly affect the mass of the system, and if so how? Will this affect the net force accelerating the system, and if so how? Since a = F_net / m, what is your conclusion about the effect on the accleration?

Ask, and answer, similar questions about the remaining systems.

Problem Number 5

A pendulum of mass .7 kg and length 3.17 meters is initially displaced .8242 cm in the horizontal direction from its equilibrium position. The pendulum is released from rest at this position.

How much work does gravity therefore do on the pendulum during its fall to the equilibrium position, and how much work does the pendulum do against gravity?
What therefore will be the change in the kinetic energy of the pendulum as it falls?
Using energy considerations determine the velocity of pendulum at its equilibrium position.

The pendulum is 3.17 meters below its fixed point when it reaches equilibrium.

How far is it below its fixed point when it is released?

How far does it therefore descend between release and equilibrium?

What therefore is its PE change as it descends?

What therefore is its KE at equilibrium?

What therefore is its velocity at equilibrium?

Problem Number 6

When masses of 45, 90 and 135 grams are hung from a certain rubber band its respective lengths are observed to be 28, 36 and 44 cm. What are the x and y components of the tension of a rubber band of length 40.31 cm if the x component of its length if 22.78722 cm?

What force directed at 149 degrees, when added to this force, will result in a horizontal net force?

What is the tension in the rubber band at each length?

Use your results to construct a graph of tension vs. length.

For the given rubber band length, use your graph to estimate the tension.

From the length and the x component of the length determine the y component of the length.

Find the angle made by the rubber band with the positive x axis.

Use this and the tension to find the x and y components of the force exerted by the rubber band.

If the new force, added to the tension, results in a horizontal net force then what must be the vertical component of that force?

You now know the vertical component and the angle of the new force. What therefore must be the magnitude of this force?

Problem Number 7

Give an example of a situation in which you are given v0, a and Ds, and reason out all possible conclusions that could be drawn from these three quantities, assuming uniform acceleration. Accompany your explanation with graphs and flow diagrams. Show how to generalize your result to obtain the symbolic expressions for vf and a.

This goes back to the Major Quiz and should be easy to answer: a and `ds are not related by either the definition of velocity or the definition of acceleration, so they don't tell you anything. v0 by itself doesn't give you any information about vAve or `dv, so it isn't can't be connected to either a or `ds by either definition. So you can't reason out anything from the definitions.

How many cm/s² of acceleration should correspond to 1 unit of ramp slope if:

a graph of number of paper clips needed to maintain equilibrium vs. ramp slope has slope 47 clips / unit of ramp slope, and
a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley has slope ( 19 cm/s²) / clip?

If we require 42 clips to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?