course Phy 121 Hey Mr. Smith, this assignment helped me to turn the corner. I hope to continue to improve. This is a very interesting (but very hard online) course. dϺe~`ܘǍgxassignment #013
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14:47:06 `q001. An object of mass 10 kg is subjected to a net force of 40 Newtons as it accelerates from rest through a distance of 20 meters. Find the final velocity of the object.
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RESPONSE --> i used the formula vf^2=V0 + 2a'ds =0 + 8m/s^2 * 20m = sq rt of 160 m =12.6 m/s confidence assessment: 2
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14:49:10 `q002. Find the value of the quantity 1/2 m v^2 at the beginning of the 20 meter displacement, the value of the same quantity at the end of this displacement, and the change in the quantity 1/2 m v^2 for this displacement.
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RESPONSE --> 1/2mv^2 at the beginning is 5 kg * 0 = 0 and at the end it would be 5 kg * 6.35 m/s = 31.75 change is 31.75 confidence assessment: 2
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14:50:22 `q003. Find the value of the quantity Fnet * `ds for the present example, and express this quantity in units of kg, meters and seconds.
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RESPONSE --> 40N * 20m = 800 kg/m^2/s^2 confidence assessment: 2
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14:50:43 `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
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RESPONSE --> they are the same confidence assessment: 3
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14:53:32 `q005. Suppose that all the quantities given in the previous problem are the same except that the initial velocity is 9 meters / second. Again calculate the final velocity, the change in (1/2 m v^2) and Fnet * `ds.
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RESPONSE --> Vf=15.5m/s initial 1/2mv^2 would be 405 final would be 1201 diff or 796 J confidence assessment: 3
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14:55:04 `q006. The quantity Fnet * `ds and the change in the quantity 1/2 m v^2 were the same in the preceding example. This might be just a coincidence of the numbers chosen, but if so we probably wouldn't be making is bigger deal about it. In any case if the numbers were just chosen at random and we obtained this sort of equality, we would be tempted to conjecture that the quantities were indeed always equal. Answer the following: How could we determine if this conjecture is correct? Hint: Let Fnet, m and `ds stand for any net force, mass and displacement and let v0 stand for any initial velocity. In terms of these symbols obtain the expression for v0 and vf, then obtain the expression for the change in the quantity1/2 m v^2. See if the result is equal to Fnet * `ds.
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RESPONSE --> the Fnet and 'ds did not change so the conjecture could not be correct,......right? confidence assessment: 2
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15:00:23 `q007. We call the quantity 1/2 m v^2 the Kinetic Energy, often abbreviated KE, of the object. We call the quantity Fnet * `ds the work done by the net force, often abbreviated here as `dWnet. Show that for a net force of 12 Newtons and a mass of 48 kg, the work done by the net force in accelerating an object from rest through a displacement of 100 meters is equal to the change in the KE of the mass.
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RESPONSE --> the a would be .25 m/s^2 vf would be 7 m/s initial KE was 0 final KE was 1200 J, a diff of 1200 'dWnet would be 12 N * 100 M or 1200 J it's the same, and thank you God, I'm starting to really get this Physics stuff ! confidence assessment: 3
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15:02:17 `q008. How much work is done by the net force when an object of mass 200 kg is accelerated from 5 m/s to 10 m/s? Find your answer without using the equations of motion.
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RESPONSE --> initial KE would be 2500 final KE would be 10000 i conclude the 'dWnet should be 7500 J confidence assessment: 3
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15:03:19 `q009. Answer the following without using the equations of uniformly accelerated motion: If the 200 kg object in the preceding problem is uniformly accelerated from 5 m/s to 10 m/s while traveling 50 meters, then what net force was acting on the object?
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RESPONSE --> it would have to be 7500 = F * 50 so the Fnet would be 150 N confidence assessment: 3
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15:06:16 `q010. Solve the following without using any of the equations of motion. A net force of 5,000 Newtons acts on an automobile of mass 2,000 kg, initially at rest, through a displacement of 80 meters, with the force always acting parallel to the direction of motion. What velocity does the automobile obtain?
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RESPONSE --> `dW=5000 * 80 = 400000 J 400000=.5mv^2 400000=1000 v^2 vf = 400 m/s i hope this is right because my confidence is rising.... confidence assessment: 3
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15:10:10 `q011. If the same net force was exerted on the same mass through the same displacement as in the previous example, but with initial velocity 15 m/s, what would then be the final velocity of the object?
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RESPONSE --> 400000= initial KE is 225000 final kE is 625000 1/2 m v^2 = 625000 1000 v^2 = 625000 v^2 = 625 vf= 25 m/s confidence assessment: 3
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15:13:34 `q012. Solve without using the equations of motion: A force of 300 Newtons is applied in the direction of motion to a 20 kg block as it slides 30 meters across a floor, starting from rest, moving against a frictional force of 100 Newtons. How much work is done by the net force, how much work is done by friction and how much work is done by the applied force? What will be the final velocity of the block?
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RESPONSE --> Fnet = 400N (300 + 100) 400 * 30 `dW= 12000 J initial KE is 0 final KE is 12000 12000 = 1/2 m v^2 vf = 34.6 m/s confidence assessment: 3
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