Query 09

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course phy121

oct 2, 4:05pm

009. `query 9

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Question: See if you can answer the following question, which came from a student:

Please define the difference between Fnet and Force.

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Your solution:

If we are using math terms, net means the sum of all. So if we say Fnet, we are talking about the sum of all the forces. Whereas force refers to only one force, for example a push or gravity or a pull. The Fnet would be the sum of all of those acting on an object.

confidence rating #$&*:

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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. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the net force exerted on the object? (It is implicitly assumed, since it is not specified otherwise, that the force and the motion of the object are in the same direction)

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Your solution:

If you are pushing an object, we would call that Work (W) and the formula for work is W= F*’ds. In order to find the force, we would need to rearrange the equation to read: F= W/’ds

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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

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?

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Your solution:

If W=F*distance, then this gives the work done, but it is general. If we change it to W=F*displacement then it is more specific. Distance is just a number where displacement is the change of position, which can be negative or positive. If the work done is positive, the object will slow down, if the work done is negative, the object will speed up.

confidence rating #$&*:

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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.

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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.

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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.

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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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `qWhy is KE change equal to the product of net force and displacement?

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Your solution:

W= F*’ds

Fnet= m*a

A= Fnet/m

If we take the 4th equation of acceleration, we can input Fnet/m in for a:

vf^2 = v0^2 + 2 a `ds.

Vf^2=v0^2+ 2*(Fnet/m)*’ds

Rearrange it to:

F `ds = 1/2 m vf^2 - 1/2 m v0^2

Therefore, KE is equal to 1/2 m vf^2 - 1/2 m v0^2.

KE0= 1/2mv0^2

KEf= 1/2mvf^2

confidence rating #$&*:

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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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: When we push an actual object with a constant force, why do we not expect that the KE change is equal to the product F * `ds of the force we exert and the distance?

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Your solution:

Since KE=F*’ds is too general, we can make it more specific to read ‘dKE= Fnet*’ds. This is the change of kinetic energy is equal to the net force multiplied by the displacement. Since there are other forces acting on the object other than the force I provide. We can say there is a friction force opposing my push, which would change the Fnet which ultimate changes the ‘dKE.

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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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: If a constant net force, exerted through a distance of 40 meters in the direction of the force, causes the kinetic energy of a 5 kg mass to change by 800 Joules then what is the net force?

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Your solution:

F= m*a

F= 5kg*9.8m/s^2

F= 49N

‘dKE= Fnet*’ds

800J= Fnet*40m

800J/40m= Fnet

Fnet= 20N

confidence rating #$&*: 1

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Self-critique (if necessary):

I found the force acting on the object, but I am not sure where that fits in with the Fnet. I guess I am just confusing myself. Would I do 49N-20N= 29N to find the overall Fnet??? Or would I put that into the equation so it would look like:

800J= (29N+F2)*40m

20N-29N= F2

F2= -9N

Therefore: 29N+(-9N)= 20N for the Fnet???

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Self-critique Rating: 2

@&

49 N is the force exerted by gravity on the object.

You don't know anything about the various forces exerted on the object. Gravity would generally be one of them, and assuming you're on or close to the surface of the Earth it would exert a 49 N force. But in most situations something else would counter the force of gravity, pushing the object in the upward direction with a 49 N force, basically canceling the effect of gravity. In that case the net force would come from something besides gravity.

In other cases, for example on an incline, you would not have a 49 N upward force, and part of the 49 N force of gravity would be in the direction down the incline, so the gravitational force could constitute part of the net force.

However on this problem none of that needs to be considered in order to answer the question.

The bottom line is just what you've said. The change in KE is the work done by the net force, and that is equal to the product of the net force and the displacement.

Your 20 N result is correct.

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Question: A constant force of 40 Newtons is applied to a certain mass as it moves 25 meters in the direction of the force. The kinetic energy of the mass increases by 750 Newtons. What net force acted on the mass? If there was another force acting, then assuming it was constant what were its magnitude and direction?

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Your solution:

KE= F*’ds

KE= 40N*25m

KE= 1000J

Fnet= 40N+750N= 810N, since it says that it increases by 750N.

The other force acting on it, would be the gravity force of 9.8m/s^2 on each kg downward, which is then providing a friction force acting in the opposite direction of the force.

confidence rating #$&*: 2

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Self-critique (if necessary):

I think I did the Fnet correctly, if not please correct.

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Self-critique Rating: 3

@&

The problem is misstated. The kinetic energy of the mass was to increase by 750 Joules.

This would imply an average force of 750 J / (25 m) = 30 N.

So there would be a -10 N force in addition to the 40 N force.

The thing that needs to be corrected is the original document, which I've just taken care of.

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&#This looks good. See my notes. Let me know if you have any questions. &#