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Note Definitions and Procedures to be Memorized

• Note for reference: Practice Major Quiz
• change default page on each computer to http://vhmthphy.vhcc.edu
• Note for reference:  Linked Outline
• Path to class notes is physics I > assts, scroll to bottom and click on Fall 04.
• Click on today's date and look at  Synopsis with Multiple Choice Question References

At h:\shares\physics run the Multiple Choice Questions choosing the qzz 'course'.  Run the following:

`m The Cart and Work-Energy

• Include comments when something isn't clear or when you think you see an error.  If you think there's an error start your comment with the word 'error'.

For homework:

Master All of Introductory Problem Set 3

Text Work:

• General College Physics:  Read Chapters 1, 2 and 4.  Do Chapter 1 problems that are multiples of 3.

• University Physics:  Read Chapters 1, 2 and 4.  Do Chapter 1 problems that are multiples of 3.

When a cart is pulled across the floor at constant velocity, the acceleration of the cart is zero so the net force on the cart is zero. The horizontal force required in the direction of motion in order to pull the cart is equal and opposite to the frictional force resisting motion.

If the pulling force is exerted wholly in the direction of motion, the work done by that force is equal to the product of the pulling force and the displacement, and is positive since the pulling force and the displacement are in the same direction.

In the meantime the frictional force is equal and opposite to the pulling force (remember that the cart is being pulled at constant velocity so that the net force must be zero; we exerted just enough pulling force to balance the friction). So the frictional force acts in a direction opposite to the direction of motion.

If the direction of motion is taken to be the positive direction, the displacement is positive and the frictional force is negative, so that the product of frictional force and displacement is negative, and the frictional force does negative work. Since the pulling force and frictional force have equal magnitudes, the work done by the frictional force is the negative of that done by the pulling force.

In symbols, if f_frict is the frictional force and F_pull the pulling force, while displacement it 15.0 , then if the positive direction is the direction of the displacement 15.0 , in the current situation F_pull and 15.0 would be positive quantities and f_frict would be negative, with F_pull = - f_frict . The work done by the pulling force is therefore `dW_pull = F_pull * 15.0 , the product of two positive quantities and therefore positive. The work done by the frictional force is `dW_frict = f_frict * 15.0 , the product of a negative quantity and a positive quantity and therefore negative. Since F_pull = - f_frict and since 15.0 is the same in both expressions we conclude that `dW_pull = - `dW_frict .

If a force of 35.0 Newtons in the direction of motion is exerted on an object which experiences frictional resistance 15.6 Newtons through displacement 27.0 then using the direction of motion as the positive direction, then the net force on the object is F_net = 35.0 N - 15.6 N = 35.0 N. The work done by the net force is then

-3460.0 = `Fnet * 27.0 = 946. N * m = 946. Joules = 946. kg m^2 / s^2.
 

If the object has mass 30 kg then what is its final velocity, assuming initial velocity 0 and `dWnet = 520 J?

We know that 520 J is the work done by the net force and is therefore change in KE.

We know that initial vel is 0 so initial KE is 0.  The change in KE therefore gives us the final KE, which is

• KEf = 0 J + 520 J = 520 J.

Knowing the KE and m we can set .5 m v^2 equal to this value and solve for v:

• .5 mv^2 = KEf.  Doing the algebra

v = +- sqrt( 2 * KEf / m) =

+_ sqrt( 2 * 520 J / (30 kg) ) =

+- 5.9 sqrt( (kg m^2 / s^2 ) / kg) =

+- 5.9 sqrt( m^2 / s^2) =

+- 5.9 m/s.

The work done by the pulling force is `dW_pull = 35.0 N * 27.0 m = 946. Joules.

The work done by the frictional force is `dW_frict = `F -frict * 27.0 = 419. Joules.

The net work is the sum of these two, or -3460.0 = 946. Joules + (419. Joules) = 946. Joules, consistent with the previous result for -3460.0 .

 If a force of 65.0 Newtons in the direction of motion is required to pull a mass of 10.0 kg across the floor at constant velocity against the frictional force, then as the mass moves a distance 15.0 meters the pulling force does work

`dW_pull = 65.0 Newtons * 15.0 meters =

976. N m =

976. Joules, or in fundamental units

976. kg m^2 / s^2.

To accomplish constant velocity the pulling force must be equal and opposite to the frictional force, which must therefore be -F_pull Newtons and which does work

`dW_frict = -F_pull Newtons * 15.0 meters = -976. Joules.

The net force on the system is F_pull + F_friction = 65.0 Newtons + (-65.0 Newtons) = 0, so the work done by the net force is zero.

Since the object moves at constant velocity, its KE .5 m v^2 does not change, so `dKE = 0.

This is consistent with the work-energy theorem, which says that `dW_net = `d(KE ).

 If a cart of mass 13.0 kg changes velocity from 27.0 m/s to 14.0 m/s as it coasts to rest against friction over a displacement of 4.60 meters, then the change in the KE of the cart is

`dKE = .5 * 13.0 kg * (14.0 m/s)^2 - .5 * 13.0 kg * (27.0 m/s)^2 = 1270.0 Joules - 4740.0 Joules = -3460.0 Joules.

By the work-energy theorem the work done by the net force is therefore -3460.0 Joules. This work is done over a distance of 4.60 meters.

It follows that the net force is the force F_net such that

F_net * displacement = -3460.0 , and that

F_net = -3460.0 / displacement =
-3460.0 Joules / (4.60 meters) =
-753. J / m =
-753. ( kg m^2 / s^2 ) / m =
-753. kg m/s^2 =
-753. Newtons.

If friction is the only force acting along the direction of motion then since normal force and gravity are in this case equal and opposite, the frictional force is the net force. Note that the frictional force is negative, hence in the direction opposite the motion.

Present version of the work-energy theorem is

• `dW_net = `dKE

This is the view from outside the system.  Using ON to designate work done on the system we can write

• `dW_netON = `dKE.

The forces exerted BY the system are equal and opposite to the forces exerted ON the system.  So if we use BY to designate work done BY the system AGAINST the forces acting on it we have

• -`dW_netBY = `dKE.

This can be rearranged to give us

• `dW_netBY + `dKE =

Interpret F vs. x graphs, including linear graph for pendulum.