query 09

.............................................

.............................................

.............................................

@&

The logic of the terminology is fairly simple:

Work done on a system is equal and opposite to work done by the system.

So if the work done on a system is positive, the work done by a system is negative.

*@

@&

As a simple example, consider pushing a car. The car is the system.

If the car is moving forward and you are pushing backward (e.g., trying to slow a car that's coasting down a slope (not necessarily a good idea)), then the force exerted by the car against your push, being equal and opposite to your push, is in the direction of motion. So the car does positive work. You, on the other hand, are pushing in the direction opposite motion, so you are doing negative work. (One effect of doing negative work is that it doesn't get you out of breath. If you exert the same push on the back of the car as it moves forward, you will in time get out of breath, provided you are pushing hard enough and the car is moving fast enough. But pushing back against the front of the forward-moving car would not get you particularly out of breath.)

If you push the car from behind as it moves forward, the car is exerting an equal and opposite force on you, so the force it exerts is in the direction opposite its motion. The car therefore does negative work, which actually means that positive work is done on it (by your force, which is in the direction of motion).

*@

.............................................

@&

The definition comes from the above analysis.

As the analysis shows,

F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

F_net _ `ds is the work done by the net force.

The right-hand side is the change in the quantity 1/2 m v^2.

We choose to give that quantity the name "kinetic energy".

Kinetic energy could be defined for a mass subject to a net force as "that which changes as a result of the work done by a net force, its change being equal to that work".

It should be clear why kinetic energy can be thought of as energy of motion.

It should become clear to you over the next few assignments why the definition and concept of kinetic energy are very useful in helping us think about the behavior of moving objects and systems subject to various forces.

*@

.............................................

@&

F_net = m * a, where m is the mass.

This is the connection to the 5 kg of this particular problem.

You use the work-energy theorem to find F_net, given the displacement and the change in KE.

Then, having the mass, you could find out the acceleration of the object, or if you wanted the change in v^2.

However neither question was asked. So for this specific problem the 5 kg is superfluous information. Not superfluous to understanding what's going on, but superfluous to answer the specific question posed here.

*@

&#Good responses. See my notes and let me know if you have questions. &#

query 09

#$&*

.............................................

Given Solution: Net force is the sum of all forces acting on an object. Typically a number of forces act on a given object. The word 'force' can be used to refer to any of these forces, but the word 'net force' refers exclusively to the sum of all the forces (for future reference note that the word 'sum' refers to a vector sum; this idea of a vector sum will be clarified later). If you're pushing your car on a level surface you are exerting a force, friction is opposing you, and the net force is the sum of the two (note that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving. The acceleration of the car depends on the net force. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:

.............................................

Given Solution: Knowing the distance `ds and the work `dW we use the basic relationship `dW = F_net * `ds Solving this equation for F we obtain F_net = `dW / `ds &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:

.............................................

Given Solution: First answer to the question (work = force * distance): This first answer serves to give you the main idea: the KE change is equal to the work done by the net force. the work done by the net force is the product of the force and the distance through which it acts so the KE change is equal to the product of the force and the distance. First answer modified to consider directions of force and motion (work = force * displacement in direction of force): The previous answer applies only if the net force is in same the direction as the motion. More correctly: the KE change is equal to the work done by the net force. the work done by the net force is the product of the force and the displacement (not 'distance') in the direction of the force the KE change is equal to the product of the force and the displacement in the direction of the force. The key difference here is the use of the word 'displacement' rather than 'distance'. Since a displacement, unlike a distance, can be positive or negative, so the work done by a force can be positive or negative. Another thing to keep in mind for the future is that the displacement is to be in the direction of the force. A negative displacement therefore denotes a displacement in the direction opposite the force. We will later encounter instances where the force is not directed along the line on which the object moves, in which case the work will be defined as the force multiplied by the component of the displacement in the direction of the force. Sometimes we will want to think in terms of the forces exerted ON objects, sometimes in terms of the forces exerted BY objects. The above statement of the work-KE theorem is in terms of the forces exerted ON an object. The basic idea is simple enough. If a force is exerted ON an object in its direction of motion, the work is positive and the object tends to speed up. On the other hand if the object exerts a force in its direction of motion, it tends to slow down. Positive work done ON an object tends to speed it up (increasing its PE), positive work done BY an object tends to slow it down (decreasing its PE). The above ideas are expanded below to consider forces exerted ON objects vs. forces exerted BY objects. Synopsis of work-kinetic energy: First be aware that because of Newton's Second Law, there are typically two equal and opposite net forces, the net force which acts on a system and the net force which is exerted by the system. It is necessary to be careful when we label our forces; it's easy to mix up forces exerted by a system with forces exerted on the system. The first basic principle is that the work by the net force acting ON the system is equal and opposite to the work done by the net force exerted BY the system. The KE, on the other hand, is purely a property OF the system. The kinetic energy change OF the system is equal to the work done by the net force acting ON the system. The kinetic energy change OF the system is therefore equal and opposite to the work done by the net force exerted BY the system. Intuitively, when work is done ON a system things speed up but when the system does work things have to slow down. A more specific statement would be If positive work is done ON a system, the total kinetic energy of the system increases. If positive work is done BY a system, the total kinetic energy of the system decreases. (We could also state that if negative work is done ON a system, its total KE decreases, which should be easy to understand. It is also the case that if a system does negative work, its total KE increases; it's easy to see that this is a logical statement but most people fine that somehow it seems a little harder to grasp). Below we use `dW_net_ON for the work done by the net force acting ON the system, and `dW_net_BY for the work done by the net force being exerted BY the system. The work-kinetic energy theorem therefore has two basic forms: The first form is `dW_net_ON = `dKE which states that the work done by the net force acting ON the system is equal to the change in the KE of the system. The second form is `dW_net_BY + `dKE = 0 which implies that when one of these quantities is positive the other is negative; thus this form tells us that when the system does positive net work its KE decreases. Summary: work = force * distance gives us the general idea but needs to be refined work = force * displacement is a correct definition as long as motion is along a straight line parallel to the force work = force * displacement in the direction of the force is true for all situations If the net force does positive work on the system, the system speeds up. Negative work on the system slows it down. More precisely: `dW_net_ON is the work done by the net force acting ON the system, and is equal to the KE change of the system. This is the work-kinetic energy theorem. One alternative way of stating the work-kinetic energy theorem: Forces exerted on the system are equal and opposite to forces exerted by the system, so If the net force exerted by the system does positive work the system slows down. Negative work done by the system speeds it up: `dW_net_BY is the work done by the net force exerted BY the system, and is equal and opposite to the KE change of the system This expanded discussion is in a separate document at the link Expanded Discussion of Work and KE . It is recommended that you bookmark this discussion and refer to it often as you sort out the ideas of work and energy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? I too am having trouble grasping the idea that negative work done BY a system speeds it up. Or, for that matter, negative work done by a system, period. Am I thinking of ""work"" too colloquially? Could you desribe a situation where a system does negative work and speeds up??? ------------------------------------------------ Self-critique Rating:

@& The logic of the terminology is fairly simple:

Work done on a system is equal and opposite to work done by the system.

So if the work done on a system is positive, the work done by a system is negative. *@

@& As a simple example, consider pushing a car. The car is the system.

If the car is moving forward and you are pushing backward (e.g., trying to slow a car that's coasting down a slope (not necessarily a good idea)), then the force exerted by the car against your push, being equal and opposite to your push, is in the direction of motion. So the car does positive work. You, on the other hand, are pushing in the direction opposite motion, so you are doing negative work. (One effect of doing negative work is that it doesn't get you out of breath. If you exert the same push on the back of the car as it moves forward, you will in time get out of breath, provided you are pushing hard enough and the car is moving fast enough. But pushing back against the front of the forward-moving car would not get you particularly out of breath.)

If you push the car from behind as it moves forward, the car is exerting an equal and opposite force on you, so the force it exerts is in the direction opposite its motion. The car therefore does negative work, which actually means that positive work is done on it (by your force, which is in the direction of motion).

*@

.............................................

Given Solution: This comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE, so that F `ds = `dKE. Here F is the net force acting on the system, so we could more specifically write this as F_net_ON = `dKE. STUDENT QUESTION: I do not see how you go from KE = 1/2 m v^2 to F_net 'ds = kEf - Ke0 INSTRUCTOR RESPONSE: If KE = 1/2 m v^2, then KEf = 1/2 m vf^2 stands for the KE at the end of the interval and KE0 = 1/2 m v0^2 stands for the KE at the beginning of the interval. Then F_net `ds = 1/2 m vf^2 - 1/2 m v0^2 becomes F_net `ds = KEf - KE0. STUDENT COMMENT In my answer I simply related it to work, I didn’t realize It was supposed to be derived from a formula. Either way, I have read through the solution and almost fully understand. I am only slightly confused by the initial choice of formula. Was this just because these were the units that were given? INSTRUCTOR RESPONSE The definition of KE can be regarded as coming from the formula. The formula is there, and when we substitute a = F / m we get quantities which we define as work and KE. The question that motivates us to do this is 'what happens when a certain force is exerted over a certain distance?' This question can be contrasted with 'what happens when a certain force is exerted over a certain time interval?'. When we answer this question, we get the quantities we define as impulse and momentum. University Physics Students Note: The formula approach outline above is based on the equations of uniformly accelerated motion. However the concept of work and kinetic energy applies whether acceleration is uniform or not. If a force F(x) is applied over a displacement interval from x_0 to x_f, we define the work to be the definite integral of F(x) with respect to x, over this interval, and it isn't difficult to show that the result is the change in the KE. If F(x) is constant, then the result is equivalent to what we get from the equations of uniform acceleration. Similarly if force F(t) is applied over a time interval, an integral leads us to the general definitions of impulse and momentum. STUDENT QUESTION I do not understand why you related this to that one specific equation INSTRUCTOR RESPONSE I assume you mean the equation vf^2 = v0^2 + 2 a `ds. The original question concerned the effect on velocity of applying a given force on a given mass through a given displacement, starting with a given initial velocity. The given force and mass imply the acceleration. Acceleration, initial velocity and acceleration imply the final velocity. So the equations arise naturally from the question. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? How do we come to define KE as 1/2 mv^2? I follow everything leading up to that step, and I see how that definition leads us to translate 1/2 m vf^2 to KEf and 1/2m v0^2 to KE0, and eventually simplify the whole thing to `dKE. But I'm not sure where that definition comes from. It looks familiar, but I don't know why. ??? ------------------------------------------------ Self-critique Rating:

@& The definition comes from the above analysis.

As the analysis shows,

F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

F_net _ `ds is the work done by the net force.

The right-hand side is the change in the quantity 1/2 m v^2.

We choose to give that quantity the name "kinetic energy".

Kinetic energy could be defined for a mass subject to a net force as "that which changes as a result of the work done by a net force, its change being equal to that work".

It should be clear why kinetic energy can be thought of as energy of motion.

It should become clear to you over the next few assignments why the definition and concept of kinetic energy are very useful in helping us think about the behavior of moving objects and systems subject to various forces. *@

.............................................

Given Solution: Change in KE is equal to the work done by the net force, not by the force I exert. i.e., `dKE = F_net * `ds The net force is not generally equal to the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:

@& F_net = m * a, where m is the mass.

This is the connection to the 5 kg of this particular problem.

You use the work-energy theorem to find F_net, given the displacement and the change in KE.

Then, having the mass, you could find out the acceleration of the object, or if you wanted the change in v^2.

However neither question was asked. So for this specific problem the 5 kg is superfluous information. Not superfluous to understanding what's going on, but superfluous to answer the specific question posed here. *@

query 09

.............................................

.............................................

.............................................

@&

The logic of the terminology is fairly simple:

Work done on a system is equal and opposite to work done by the system.

So if the work done on a system is positive, the work done by a system is negative.

*@

@&

As a simple example, consider pushing a car. The car is the system.

If the car is moving forward and you are pushing backward (e.g., trying to slow a car that's coasting down a slope (not necessarily a good idea)), then the force exerted by the car against your push, being equal and opposite to your push, is in the direction of motion. So the car does positive work. You, on the other hand, are pushing in the direction opposite motion, so you are doing negative work. (One effect of doing negative work is that it doesn't get you out of breath. If you exert the same push on the back of the car as it moves forward, you will in time get out of breath, provided you are pushing hard enough and the car is moving fast enough. But pushing back against the front of the forward-moving car would not get you particularly out of breath.)

If you push the car from behind as it moves forward, the car is exerting an equal and opposite force on you, so the force it exerts is in the direction opposite its motion. The car therefore does negative work, which actually means that positive work is done on it (by your force, which is in the direction of motion).

*@

.............................................

@&

The definition comes from the above analysis.

As the analysis shows,

F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

F_net _ `ds is the work done by the net force.

The right-hand side is the change in the quantity 1/2 m v^2.

We choose to give that quantity the name "kinetic energy".

Kinetic energy could be defined for a mass subject to a net force as "that which changes as a result of the work done by a net force, its change being equal to that work".

It should be clear why kinetic energy can be thought of as energy of motion.

It should become clear to you over the next few assignments why the definition and concept of kinetic energy are very useful in helping us think about the behavior of moving objects and systems subject to various forces.

*@

.............................................

@&

F_net = m * a, where m is the mass.

This is the connection to the 5 kg of this particular problem.

You use the work-energy theorem to find F_net, given the displacement and the change in KE.

Then, having the mass, you could find out the acceleration of the object, or if you wanted the change in v^2.

However neither question was asked. So for this specific problem the 5 kg is superfluous information. Not superfluous to understanding what's going on, but superfluous to answer the specific question posed here.

*@

&#Good responses. See my notes and let me know if you have questions. &#