Work, Energy and Momentum Notes 1 – Work

Work is defined as the transfer of energy from one body to another. Or more rigorously:

We can calculate the work done on an object with:

the units of work are Nm or Joules Note that these are the same units as torque yet their values are used to describe very different quantities.

Although a seemingly simple idea, the concept of work is often misunderstood. Let’s look at a few examples to help clarify.

Example 1 - Work against Gravity How much work is required to lift a 2.0 kg textbook from the floor to a height of 1.5 m at a constant velocity? Note: W = Fd, but what force do we need to exert to lift the book at a constant velocity? Since the velocity is constant what is the net force acting on the book?

Example 2 - Work on an object How much work is done on a 4.0 kg medicine ball that is held at a height of 1.8 m for 10 s? Note: Is energy being used to hold the ball in this position? Is work actually being done ON THE BALL?

Example 3 – Forces at an angle The plucky youngster pictured to the right is pulling his sled at a constant velocity of 1.2 m/s. He pulls the 15 kg sled with a force of 35 N at an angle of 40o to the horizontal. How much work does he do in pulling the sled 20 m? Note: Draw an FBD showing the forces at work on the sled. Break Fboy into its vertical and horizontal components. Does the vertical component of the force do any work?

Rule: When finding the work done on an object we only consider…

Example 4 – Fnet vs. Fapp A biology student is pushing a rope up a hill. The student pushes the rope with a force of 220 N while the force of friction is 120 N. How much work is the student doing? Note: To find the amount of work done by the student should we used Fnet or Fapp?

Rule: When finding the total work done by a body we always use:

Hey wait a minute, if work is the transfer of energy what the heck kind of energy is generated by the force of friction?

Rule: When finding the amount of energy is lost due to friction we use:

Example 5 – To scalar or not to scalar? Work is the product of two vectors so (of course) it is a ______________. However work can be positive or negative…but how? Glad you asked. Imagine that you bring a 1.0 kg basketball from the floor to the top of a 1.0 m table. How much work did you do? Wow, you’re awesome. Which direction did you exert that force, Pipes? Now suppose the ball rolls off the table and falls straight down to the floor. How much work was done on the ball? Be careful: which direction is the force working on the ball now? Rule: Work can be negative when the force doing the work acts in the negative direction. Another way of thinking of this is to remember that work is a __________ ____________________. When you pick the ball up off the floor you are actually transferring energy to the ball in the form of _________________ energy. When the ball falls off the table, it is losing that energy.

Work, Energy and Momentum Notes 2 – Potential and Kinetic Energy

There are many forms of energy: mechanical, thermal, electrical, nuclear, chemical etc. One form can be converted into another by doing work. In this chapter will be concerned mostly with potential and kinetic (and just a hint of thermal) energy.

Potential Energy (Ep):

Ep = Fd

For gravitational potential energy:

Ep =

Example A 1000 kg boulder sits on a 50 m ledge, precariously perched above a biology student. How much potential energy does the boulder have relative to the student?

Cause math is fun! Deriving the Ep formula…

Ep = Fd

(in this case F = Fg = mg)

Ep = mgd (let’s change letters just for fun,

and call h the vertical displacement)

Ep = mgh

HOORAY!

Remember: Potential energy is always…

Kinetic Energy (Ek): Ek =

Example Remember that biology student hanging out below the 1000 kg boulder? If the boulder falls the full 50 m onto the student below, how much kinetic energy will it have just before impact?

Cause math is fun! Deriving the Ek formula…

v2 = vo

2 + 2ad (take vo = 0)

v2 = 2ad (but a = F/m)

v2 = 2Fd m

(but Fd=W= Ek, when vo = 0)

v2 = 2Ek m

Ek = 1/2mv2

WA WA WEE WA!!

Work-Energy Theorem for Net Force One last thing on kinetic energy… It is important to notice that the work done by the net force on an object is equal to the change in its kinetic energy:

Wnet = ∆Ek

Or

Fnetd = 1/2m(v2 – vo2)

Note: In this case we use Fnet because …

Work, Energy and Momentum Notes 3 – The Law of Conservation of Energy

Energy cannot be created or destroyed, only changed from one form into another. Therefore in a closed system the _________ energy is always _____________.

When only conservative forces (such as gravity) act on an object kinetic energy is transferred into potential energy and vice versa. Consider a ball being thrown up into the air and returning to the thrower.

In this case The Law of Conservation of Energy states that the total amount of energy is constant, or the total change in energy is ZERO:

Total Energy = ∆Ek + ∆Ep = 0

1/2m∆v2 = mg∆h

Gain in Ek = Loss in ∆Ep

Example A ball is thrown in the air with a velocity of 14 m/s. How high is it when it has a velocity of 4.0 m/s?

Note:

Another way of thinking of The Law of Conservation of Energy is that in a closed system the total energy must be constant.

Or the total initial energy must equal the total final energy.

Total Energy = ∆Ek + ∆Ep = 0

1/2m(v2 – vo2) = - mg(hf - hi)

Example The first peak of a roller coaster is 55 m above the ground. The 1200kg car starts from rest and goes down the hill and up the second hill which is 30 m high. How fast is the car traveling at the top of the second hill?

Insert excellent picture here.

Back in grade 11 it really was that easy… When non-conservative forces (such as friction) act on an object, not all energy is transferred between kinetic and potential. This is what physicists have termed REALITY. Deal with it. The “work” done by friction does produce another form of energy known as:

______________ or _______________________ This energy is quickly conducted or radiated in all directions and effectively dispersed. Consider a block of wood sliding down a ramp with a small amount of friction.

How would the block’s kinetic energy at the bottom compare to its potential energy at the top? Why?

The fact that the amount of energy in the block decreases as it slides down the ramp doesn’t change the fact that the ________ __________ in the system is CONSTANT. We need modify our earlier equation for The Law of Conservation of Energy only slightly:

Total Energy Initial = Total Energy Final

Example A 5.0 kg block of wood is now pushed down a ramp with a velocity of 6.0 m/s. At the bottom of the ramp it is traveling at 7.5 m/s. How much thermal energy is generated due to friction?

1.5 m

3.5 m

A little Crippler for you..

Work, Energy and Momentum Notes 4 – Power and Efficiency

In everyday language we often use the words WORK, ENERGY and POWER synonymously. However this makes the physics gods extremely furious because we should all know that: POWER is…

This is extremely important in real world applications. For example Mr Trask can often be heard bragging that his 1991 Turbo Junkmobile can easily do 100 BILLION joules of work. Is he telling the truth?

Why is this still unimpressive?

Mathematically we define power as:

P = =

The unit of power is J/s or Watts (W) (this is sometimes confusing because W is also the symbol for work)

Example: A physics student is setting up a wicked body slam on a biology student. He lifts the 75 kg student clear over his head to a height of 2.2 m in 0.675 s. How much power did the physics student generate?

Example: While cruising along in the CT-2004 at 12.0 m/s, Mr Trask hits the gas and speeds up to 26.0 m/s in 1.5 s. If the CT-2004 weighs 6.50 x 104 N how much power did it generate? Ignore friction.

Another useful equation for power can be derived:

P = W/t

Example: A student pushes 14 kg of their physics homework up a 40o ramp at a constant velocity of 3.2 m/s. The friction force is 26 N. How much power must the student exert?

Whenever we use a machine to do work some of the energy we put into the machine is always lost, mainly due to friction. For example an electric heater is ________ efficient

a car is __________________ efficient a lightbulb is _____________ efficient

We can define efficiency in two ways:

Efficiency = =

Notice that efficiency is a ratio expressed as a percentage and therefore has no units!

The most common source of confusion when calculating efficiency is in understanding which values applies to work/power IN and which applies to work/power OUT. Work/Power In: Work/Power Out: Remember that energy is always LOST somewhere in using the machine, so

Work IN Work OUT

and

Efficiency

Example: A 300 W electric motor lifts a 25.0 kg object to a height of 10.0 m in 11.5 s. What is the efficiency of the motor?

Example: A student pushes a 15 kg box up a ramp with a force of 95 N. What is their efficiency? 5.0 m

9.0 m

Work, Energy and Momentum Notes 5 – Momentum and Collisions

Momentum is a quantity of motion that depends on both the mass and velocity of the object in question.

∆p =

The units of momentum are therefore ___________or ___________.

Momentum is a _________ quantity, with the same sign as its velocity. As with any vector you can assign any direction as positive and the opposite as negative, but as convention we will generally refer to up or to the right as positive and down or to the left as negative.

Momentum is similar to inertia - the tendency of an object to remain at a constant velocity. Where as inertia depends only on mass, momentum depends on mass AND velocity.

Consider this: A tennis ball traveling at 10 m/s and a medicine ball traveling at 10 m/s. Which one is more difficult to stop? OK now this one: A baseball traveling at 5 m/s and a baseball traveling at 50 m/s. Which one is harder to stop?

Example: A 350 g egg is dropped from a height of 7.5 m and splatters on the floor. Determine the change in momentum of the egg during the splatter. Assume the post splatter velocity to be zero.

Example: A baseball pitcher hurls a ball at 32 m/s. The batter crushes it and the ball leaves the bat at 48 m/s. What was the ball’s change in momentum?

Remember that momentum is a VECTOR which means:

The Law of Conservation of Momentum Momentum is a useful quantity because in a closed system it is always conserved. This means that in any collision, the total momentum before the collision must equal the total momentum after the collision.

pinitial = pfinal

for a collision involving a total of n bodies,

First we are going to consider linear interaction – essentially 1-D There are 3 general types of these collisions: (1) Collision where the two bodies don’t stick together

Example: A 1200 kg car heading east at 35 m/s collides with a 1550 kg truck traveling west at 27 m/s. After the collision the truck is traveling east at 6.0 m/s. What is the final velocity of the car?

Before After m1i = m2i = m1f = m2f = v1i = v2i = v1f = v2f =

(2) Collision where the bodies do stick together Example: A 9500 kg caboose is at rest on some tracks. An 1100 kg engine moving east at 12.0 m/s collides with it and they stick together. What is the velocity of the train cars after the collisions?

Before After m1i = m2i = mtotal = v1i = v2i = vf =

(3) Explosion Example: An object…no wait…a fluffy little bunny suddenly explodes in a stunning display of gore. The bunny splits into exactly two parts, the first part has a mass of 2.2 kg and flies due east at 26 m/s. The second chunk heads due west at 34 m/s. What was the initial mass of the bunny?

Before After

mtotal = m1f = m2f =

vi = v1f = v2f =

Work, Energy and Momentum Notes 6 – Impulse

Whenever a net force acts on a body, an acceleration results and so its momentum must change. We define the change in an object’s momentum as _________________. Recall that momentum is the product of _________ and ___________. Since we will not be dealing with changing masses, we can define an object’s change in momentum as:

∆p = m∆v

Keep that in mind while we do a little deriving… If Fnet = ma and a = ∆v/t So our equation for impulse can be written:

But what does this all mean? Let’s try to understand how forces relate to changes in momentum with a few examples. Example Coaches for many sports such as baseball, tennis and golf can often be heard telling their athletes to “follow through” with their swing. Why is this so important?

Example A student jumps off a desk. When they land they bend their knees on impact. Why does this help prevent some serious damage to their knees? Example Boxers are taught to snap their jabs out and back in as fast as possible. Why would this be advantageous when trying to apply force to their opponents face? Example A beanbag and a high bounce ball are dropped from the same height. Which one exerts a greater average force on the floor?

Example A net force of 14.0 N acts south on a 6.00 kg object for 0.10 s. If the object was initially travelling at 4.0 m/s north what is the velocity of the object?

Alright let’s do some math! Example A net force of 12.0 N north acts on an object for 2.00 x 10-3 s. Calculate the impulse.

Work, Energy and Momentum Notes 7 – Elastic and Inelastic Collisions

Collisions can be grouped into two categories, elastic and inelastic. Elastic Collisions: Inelastic Collisions:

In reality collisions are generally somewhere in between perfectly elastic and perfectly inelastic. As a matter of fact, it is impossible for a macroscopic collision to ever be perfectly elastic. Perfectly elastic collisions can only occur at the atomic or subatomic level. Why can’t macroscopic collision ever be elastic? However if objects collide and don’t stick together then there is some elastic component of the collision.

Example A 225 g ball moving at 30.0 m/s to the right collides with a 125 g ball moving in the same direction at a velocity of 10.0 m/s. After the collision the velocity of the 125 g ball is 24.0 m/s to the right. a. What is the velocity of the 225 g ball after the collision? Before After b. Is this collision elastic or inelastic? Prove it mathematically. c. What happened to the kinetic energy lost?

Work, Energy and Momentum Notes 8 – Collisions in 2-D

When dealing with collisions in 2-dimensions it is important to remember that momentum is a vector with magnitude and direction. When finding the ____________ _________________, we have to do ________________ ____________________. Collisions at 90o: A 4.0 kg object is traveling south at a velocity of 2.8 m/s when it collides with a 6.0 kg object traveling at a velocity of 3.0 m/s east. If these two objects stick together upon collision, at what velocity do the combined masses move? m1 = 4.0 kg m2 = 6.0 kg mtotal = _________ v1 = 2.8 m/s v2 = 3.0 m/s vtotal = _________ p1 = _______ p2 = _______ ptotal = _________ To find the total momentum, add the two initial vectors and find the resultant: Remember that it is momentum that is conserved, so we need to add the _________________ ______________ NOT velocity!!!

1 1+2 2

Before After

Solve for the resultant: From the resultant momentum find the final velocity (magnitude and direction): Collisions not at 90o (because life is never that easy…): A 4.0 kg bowling ball is moving east at an unknown velocity when it collides with a 6.1 kg frozen cantaloupe at rest. After the collision, the bowling ball is traveling at a velocity of 2.8 m/s 32o N of E and the cantaloupe is traveling at a velocity of 1.5 m/s 41o S of E. What was the initial velocity of the bowling ball? m1 = m2 = m1 = m2 = v1 = v2 = v1 = v2 = p1 = p2 = p1 = p2 =

Before After

1 1

2

2

There are two ways to solve this problem - the scalpel or the sledgehammer. Component Method a.k.a. The Scalpel (for the discerning physicist): We need to break the final momenta of the two objects into x and y components: We then add the individual x and the individualy components to find our total momentum. Σpx = p1x + p2x = Σpy = p1y + p2y = Notice that the total momentum is all in the ___ ______________! This should be no surprise since the bowling ball was initially only moving in the x direction. Don’t forget to solve for the initial velocity (magnitude and direction):

p1

p1x =

p1y = p2

p2y =

p2x =

The Sledgehammer aka Vector Addition (because hammers are fun): To solve this problem we simply add the vectors and solve for c with the cosine law. Notice that the __________ ______________ is either the initial or the final because momentum is ___________________. First we need to use geometry to solve for the angle opposite the total momentum. And then, start hammering:

p1 p2

ptotal