course Phy 201 Here is my work for Asst 15. I
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17:39:11 `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 --> First we know that Fnet = 10 Newtons, mass = 2 kg and the 'dt = 3 s. We can solve for the acceleration by : a = 10 Newtons / 2 kg = 5 m/s^2 Then with the acceleration we can then find the 'dv: a = 'dv / 'dt 'dv = 5 m/s^2 * 3 s 'dv = 15 m/s
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17:39:52 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 --> It took me sometime to remember how to do this one and it was so simple.
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17:44:18 `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 --> The quantity m * v changed to m = 2 kg and v = 15 m/s so that would be 20 Newton meters. As for the Fnet * 'dt, the Fnet = 10 Newtons and the 'dt = 3 s so the product of that would be 30 Joules. The two quantities are relative to one another because Joules = Newton meters.
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17:44:33 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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17:47:41 `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 --> The impulse of the vehicle will be Fnet * 'dt. So the product of this equation is Fnet = 2000 N and 'dt = 1.5 s so it is 3000 impuls of the force. I am not sure if you still call that Joules.
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17:48:30 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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18:02:30 `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 --> First we have to get the acceleration: Fnet = m * a a = 2000 N / 1200 kg = 1.666666667 m/s^2 Now the change in velocity: a = 'dv / 'dt 'dv = 1.666666667 m/s^2 * 1.5 s = 2.5 m/s
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18:03:49 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 knew I could answer it that way, but I thought you wanted me to go the other way around since you stated the question in such a way. I do understand both ways, just so you know.
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18:10:09 `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 --> Well we know that v0 = 20 m/s and vf = 25 m/s so that would mean that the change in velocity will be 'dv = 5 m/s. Now we can solve the momentum of the vehicle be m * v: 1600 kg * 5 m/s = 8000 Newton meters Since we know that the 'dt = 2 s and that the momentum of an object is equal to its impulse then we can simply plug in the values and solve for the force: Impulse = Fnet * 'dt Fnet = 8000 Joules / 2 s Fnet = 4000 Newtons
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18:10:59 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 --> Ok great for the first time in a while I really understand something about this stuff!
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T{eswarlx assignment #015 yE܃֭ Physics I 12-01-2005
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21:46:57 Set 4 probs 1-7 If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?
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RESPONSE --> We can find the change in velocity. Fnet * 'dt gives us the impulse of the object which is equal to the momentum which is equal to mass * 'dv. So substituting the impulse for the momentum and solving for 'dv will give you the answer.
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21:49:49 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
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RESPONSE --> Ok well when I think of a change in something I am trained to the thought of velocity. I guess when I read it I automatically went to that conclusion. However I feel that I explained in enough detail that you see what I was getting at. Although I did get the impulse and the momentum mixed up.
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21:50:43 What is the definition of the momentum of an object?
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RESPONSE --> The amount of force that is exerted on an object over a given time period.
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21:52:22 ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **
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RESPONSE --> Now in the last answer you said that momentum was Fnet * 'dt and that is what I thought from the lessons, but now I see that you say that it is mass * velocity. Now I know that they are both equal but what is paired with what?
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21:54:23 How do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?
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RESPONSE --> Well you would multiply the Fave by the 'dt and that would give you the answer or the equivalent to it ie. the impulse.
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21:55:29 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
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RESPONSE --> Ok well that is two in a row so I will assume that the answer in the beginning was a type O. If I am wrong let me know.
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22:00:46 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
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RESPONSE --> Newton's Second Law states that Fnet = m * a. The impulse-momentum theorem says the momentum = m * 'dv and that impulse = Fnet * 'dt. So you could say that momentum = m * ( vf - v0 ) and impulse = (m * a ) * 'dt. They are all relative to one another.
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22:03:10 ** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt **
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RESPONSE --> Ok yours looks a whole lot better than mine did. I see exactly what you did I just have a little trouble setting it up in my head at first.
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22:15:09 If you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies?
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RESPONSE --> Well you could take the mass and multiply the difference of the final and initial velocities ( 'dv). This would give you the change in momentum which is equal to the impulse. You could then take the value and set up the problem then solve for fnet: Momentum = mass * ( vf - v0 ) = ? kg m/s Impulse = Momentum Fnet = Impulse / 'dt The second would be to take the 'dv and divide it by the 'dt to give you the acceleration and then multiply it by the mass to give you the Fnet: 'dv = vf - v0 a = 'dv / 'dt Fnet = mass * a
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22:15:20 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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22:16:16 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
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22:16:20 ** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **
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