course Phy 201 ˡJgy|ƒέassignment #015
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18:46:56 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
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RESPONSE --> Fnet=10N m=2kg 'dt=3seconds vo=0 a=Fnet/mass =10N/2kg =5m/s^2 5m/s^2=vf/3seconds 15 meters/second=vf vf-vo=change in velocity 15-0=change in velocity confidence assessment: 2
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18:47:15 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
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RESPONSE --> I understand. self critique assessment: 3
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18:49:29 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
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RESPONSE --> Initially, mass*velocity was 2kg*0m/s=0 Then, the velocity changed and m*v was 2kg*15m/s=30 Fnet*'dt=10N*3seconds=30 They are the same. confidence assessment: 3'
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18:50:17 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
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RESPONSE --> I was not sure what units to put even though I carried the units through. I wanted to put the kg m/s, but I thought that ""looked"" wrong. I will try to trust myself more next time.
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18:51:52 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
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RESPONSE --> Fnet*'dt=2000N*1.5 seconds=3000Newtons/second confidence assessment: 2
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18:52:25 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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RESPONSE --> I understand. self critique assessment: 3
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18:55:18 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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RESPONSE --> Fnet=2000N m=1200kg 'dt=1.5 seconds vo=0 a=Fnet/mass =2000N/1200kg =1.7 meters/second^2 1.7m/s^2=vf/1.5 2.55=vf vf-vo=2.55-0=2.55 m/s confidence assessment: 3
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18:55:49 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
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RESPONSE --> I did it the ""old"" way. Is that okay? I still got the right answer.
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18:57:56 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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RESPONSE --> Fnet*'dt Since we assume the vehicle starts from rest, we will compare the answers at 0 seconds and at 2 seconds 1600kg*0seconds=0 kg meter/second 1600*2 seconds=3200 kg meter/second So, the change, I am guessing, will be 3200 kg meter/second. confidence assessment: 3
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18:59:30 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE --> I started out trying to find the change without first finding the change per second. Having seen this example, I feel like I could do this first step now and then complete the rest of the problem similar to what I did in my own answer. self critique assessment: 2
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