WORK.......

In physics, unlike life, work has a very specific meaning. When a force is applied to an object, and that object moves, work is done.

So if you’re pushing on a refrigerator and it never moves, you’ve done no work. Tough luck, no?

(This also means that if you sit on your...in your chair, no work is being done because no motion is involved)

We define work as (force x displacement), or

W = F(Dx)

= (Newtons) x (meters)

= Nm

= Joules (J)

giving us a new unit of measurement, the Joule.

Key Features of Work:  (these will be illustrated below)

1. Positive work is done (by the force) if the force is in the same direction as the motion.

2. No work is done by the force if it is perpendicular to the motion.

3. No work is done by any force if the object is motionless.

4. Negative work is done (by the force) if the force acts in the direction opposite to the motion.

Key Feature #1: Our definition also means that if you apply a force at an angle to the motion, only part of the force is doing work.

We can calculate the component of force responsible for work:

W = Fcosq(Dx)

Key Feature #2: Furthermore, if you apply a force to an object that is moving and the force you apply is perpendicular to the motion then that force does zero work!

An example of this would be moving a box along the floor. You do work, but the gravitational field of the earth does not.  This makes sense because the gravitational field is not contributing to the motion.  The formula for work agrees with this:

W = Fcosq(Dx)

W = Fcos90^o(Dx) = 0 !!    (the cosine of 90^o is zero)

Key Feature #3:  No motion means  (Dx) = 0.  Therefore, W = 0.

Key Feature #4: Finally, if the object is moving against a force then the work done by that force is negative. (Friction is a good example of this: it always opposes motion).

Example 1:

A box is pushed 4.0 meters across a horizontal surface with a horizontal force of 40 Newtons. A frictional force of 3.0 Newtons opposes the motion. What work was done?

Work done by the 40 newton force:

W = F(Dx)

W = (40 N)(4.0 m)

W = 160 Joules

Work done by the frictional force:

W_f = F(Dx)

W_f = (-3.0 N)(4.0 m)

W_f = -12 Joules

So the net work done on the object is the sum of these individual works:

Net Work = W + W_f

= 160 - 12 Joules

= 148 Joules

Example 2:

The same (8.0 kilogram) object is pushed with a force of 40 Newtons across a surface with a coefficient of kinetic friction, m_k = 0.20. Find the frictional work and the net work.

first you have to find the force due to friction!

Recall the equation for kinetic friction (yeh, right...)

f_k = (m_k)(F_n)

= (0.20)(mass x gravity)

= (0.20)(8.0 kg x 9.80 m/s²)

= 15.7 Newtons

and then follow through as we did above...

Energy.......

 If we take one of our old familiar motion equations, (please say it's familiar)

v_f² = v_i² + 2a(Dx)

and multiply it on both sides by mass (m), and divide both sides by 2,

(1/2)mv_f² = (1/2)mv_i² + ma(Dx)

We see that "ma" is simply the Force, or F, and we can rearrange terms:

(1/2)mv_f² - (1/2)mv_i² = F(Dx)

The terms (1/2)mv² we define as our kinetic energy, where ‘f’ indicates final, and ‘i’ indicates initial.

Notice that Work pops up! This equation can be read as follows:

Or,

KE_f - KE_i = Work

Work = (DKE)

In Example 1 above we saw that the net work was 148 Joules. Now we see that those 148 Joules will cause an increase in the energy of the "system" which is really saying the velocity of the box will increase. So, for example, if you want your car to go faster you must do work on it: (you step on the gas) the engine does work by burning fuel and that work increases the energy of your car...meaning the velocity increases (actually you're still sitting there doing no work, as usual. The engine is doing it for you)

Example 3:

A 1000 kilogram car is moving at 5 meters/sec. 40,000 Joules of work is done by the engine. What is the new speed of the car?

(1/2)mv_f² - (1/2)mv_i² = 40000 Joules

(1/2)(1000 kg)v_f² - (1/2)(1000 kg)(5 m/s)² = 40000 J

v_f = 10.2 m/s

_________________________

Sometimes the work done on a system doesn’t result in a change in kinetic energy but the work can be stored. This is called potential energy. For instance, compress a spring. The spring can then "return" the stored energy when it is released (it had the ‘potential’ to do work). Another example is the gravitational field of the Earth. If you lift an object up, you’ve just stored energy in earth’s gravitational field. How much? Let's see...

Since,

W = F(Dx)

we recognize that the "force" we have to apply is just the weight of the object. Therefore,

W = mg(Dh)

where h = height we’ve lifted the object. This equation expresses the form for gravitational potential energy:

PE_grav = mg(-Dh)   (we'll see why it's negative later)

Example 4:

How much work is necessary to lift a 80.0 kg barbell.2.50 meters? What is it’s potential energy?

W = (80.0 kg)(9.80 m/s²)(2.50 m)

W = 1960 Joules

Now this tells us how much work we have to do. The earth’s gravitational field is pointing opposite to our barbell’s motion so it is doing negative work!

W_gravity = -1960 Joules

This means that we have stored work, or energy, in the gravitational field and 1960 Joules is referred to as potential energy: because it has the potential to be returned...when the barbell is dropped on the trainer’s foot.

We can expand on this equation for Potential energy and work as follows:

W_gravity = - mg(Dh) = - mg(h-h_o) (we’re storing energy)

W_gravity = - (mgh - mgh_o)

W_gravity = - (PE_final - PE_initial)

W_gravity = - (DPE)

Notice that it is negative (because an increase in height results in negative work done by the gravitational field).

So we have three equations involving work...

1. W = F(Dx)

2. W = DKE

3. W = -(DPE)

What do you think we’ll do with these next?

If you said combine 2 & 3, let me know and I’ll be very impressed (but it still won’t get you any extra credit). Since Work is equal to both the change in Kinetic Energy and the change in Potential Energy, we can equate them,

DKE = -(DPE)

expanding, this becomes,

KE_f - KE_i = - (PE_f - PE_i )

we can rearrange these terms to get our first conservation equation:

KE_i + PE_i = KE_f + PE_f

This is called the Conservation of Mechanical Energy equation (because it doesn’t take into account heat, or dissipative, energies). It says,

Example:

A roller coaster starts at point A at a height of 90 meters and a velocity of 0.00 m/s. Find it’s speed at the bottom of the loop (B) and at the top of the next rise (C). Use the diagram below and consider it frictionless.

Since there’s no friction involved, we can use our new conservation equation...

KE_i + PE_i = KE_f + PE_f

to use this we must first substitute in expressions for the kinetic and potential energies...we note that the potential energy is gravitational potential energy (mgh):

(1/2)mv_A² + mgh_A = (1/2)mv_B² + mgh_B

Notice that I’ve used subscripts of A and B to represent the energies at these locations. Note also that this equation says: "the energy at A is equal to the energy at point B". Since the mass on both sides will cancel, we find...

(1/2)(0 m/s)² + (9.80 m/s²)(90 m) = (1/2)v_B² + (9.80 m/s²)(0 m)

v_B = 42 m/s

For the next hill we can simply change the subscripts and write:

(1/2)mv_B² + mgh_B = (1/2)mv_C² + mgh_C

Now substitute in the values for v_B, h_B, h_C, and solve for v_C. I’ll let you guys fly solo on this one. Note that we could have used these subscripts as well:

(1/2)mv_A² + mgh_A = (1/2)mv_C² + mgh_C

Because the energy at A and C must be the same also. In fact, the energy everywhere is always the same, even halfway up one of the hills: the only difference is how much is kinetic and how much is potential...

Now, finally, Conservation of Energy. This is the more general form of the equation and includes dissipative (frictional) forces: energy that is lost. The total mechanical energy we looked at above now changes at a later time or position because energy was lost to friction.

We can write this as...

KE_i + PE_i + W_fr = KE_f + PE_f

where W_fr = work done by friction. We discussed earlier that work done by friction is negative, so that this "work" lowers our initial total energy and we’re left with whatever’s on the right side of the equation (if we substitute F(Dx) for W_fr, frictional force F is negative supplying us with the desired negative sign):

Example:

A 70 kg diver leaps 10 meters into the pool and comes to rest 4 meters below the surface. Neglecting air resistance, how much energy was absorbed by the water?

We can use Conservation of Energy. He starts with no velocity (so KE_initial = 0), and we can set the lowest level to zero height (-4 also would work) so that Pe_final = 0 and since he has no remaining velocity, Ke_final also = 0!!.

KE_i + PE_i + W_fr = KE_f + PE_f

0 + mgh + W_fr = 0 + 0

(70 kg)(9.80 m/s²)(14 m) + W_fr = 0

W_fr = - 9604 Joules

 We can further extract the average frictional force through the water from the event by noting that friction only acts over 4 meters.

W = F(Dx)

-9604 J = F (4 m)

F = 2401 N