course Phy 121 009. `query 9 *********************************************
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the net force exerted on the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet=(Wd)/x Confidence rating #$&*: 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): Ok ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE=Fnet*dx confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhy is KE change equal to the product of net force and displacement? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because Fdx=1/2mv^2 and KE change=1/2mv^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self- Critique rating #$&* ********************************************* Question: When we push an actual object with a constant force, why do we not expect that the KE change is equal to the product F * `ds of the force we exert and the distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE change=work done by the net force, nothing else. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if necessary): OK Self-critique rating #$&* OK "