course Phy 231 7/9 around 8 009. `query 9
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Given Solution: Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting 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 accel of the car depends on the net force. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we know those tow things then the force is just the work divided by the distance traveled. Confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The KE is equal to the work done so it is equal to the force times the distance traveled. Confidence rating:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A slightly over-simplified simple answer: _. 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. That answer is a bit over-simplified because it 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. 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 in the direction of the force. We will later encounter instances where the force is not directed along the line on which the object moves. These ideas are expanded below. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not understand that it matters on the direction of the motion. Self-critique Rating:2 ********************************************* Question: `qWhy is KE change equal to the product of net force and displacement? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We find this by looking at the equation vf^2 = v0^2 + 2 a `ds. So since a = Fnet / m. Then we can say vf^2 = v0^2 + 2 Fnet / m `ds. Since we can say that KE is 1/2 m v^2 this makes the above equation F `ds = `dKE. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to add the KE0 to the final equation but now I know that and why it is there. Self-critique Rating: 2 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because in the real world we have other forces acting back on the force like friction and wind resistance. These will lower the KE of the system making it not equal to the net force. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: "