Asst 13

Good.

If anything is not clear let me know, and include as many specifics as possible.

You're doing well, picking up the main ideas and applying them. I've inserted several notes. Let me know if you have questions.

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19:39:01 `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 --> We know the mass = 10 kg, Fnet = 40 Newtons, 'ds = 20 meters and the v0 = 0 cm/s. We can find the acceleration by : a = Fnet / mass a = 40 Newtons / 10 kg a = 4 m/s^2 We can find final velocity by : vf^2 = v0^2 + 2a'ds vf = sqrt ( ( v0 )^2 + 2 ( 4 m/s^2)( 20 m ) vf = sqrt ( 2 ( 80 m^2/s^2 )) = sqrt ( 160 m^2/s^2) vf = 12.65 cm/s

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19:39:19 We know the initial velocity v0 = 0 and the displacement `ds = 20 meters. We have the information we need to determine the acceleration of the object. Once we find that acceleration we can easily determine its final velocity vf. We first find the acceleration. The object is subjected to a net force of 40 Newtons and has mass 10 kg, so that will have acceleration a = Fnet / m = 40 Newtons / 10 kg = 4 m/s^2. We can use the equation vf^2 = v0^2 + 2 a `ds to see that vf = +- `sqrt( v0^2 + 2 a `ds ) = +- `sqrt ( 0 + 2 * 4 m/s^2 * 20 meters) = +-`sqrt(160 m^2 / s^2) = +-12.7 m/s. The acceleration and displacement have been taken to be positive, so the final velocity will also be positive and we see that vf = + 12.7 m/s.

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19:47:43 `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 --> At the beginning the problem would be: KE = .5 ( 10 kg ) ( 0 cm/s )^2 KE = 5 Joules At the end it would be: KE = .5 ( 10 kg ) ( 12.7 cm/s ) KE = 63.5 Joules The change in the quantity would be 58.5 Joules positive.

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19:48:50 Over the 20 meter displacement the velocity changes from v0 = 0 m/s to vf = 12.7 m/s. Thus the quantity 1/2 m v^2 changes from initial value 1/2 (10 kg) (0 m/s)^2 = 0 to final value 1/2 (10 kg)(12.7 m/s)^2 = 800 kg m^2 / s^2.

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RESPONSE --> I made a big booboo in dropping the sqrt of 2. And boy does that change the outcome.

It does make a difference to square the velocity--note that it also makes the units work out right. However except for that you had the procedure right, which is very good.

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19:52:07 `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 --> Fnet * 'ds can be expressed as: 40 Newtons * 20 kg = 800 kg m^2/s^2

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19:53:22 Fnet = 40 Newtons and `ds = 20 meters, so Fnet * `ds = 40 Newtons * 20 meters = 800 Newton meters. Recall that a Newton (being obtained by multiplying mass in kg by acceleration in m/s^2) is a kg * m/s^2, so that the 800 Newton meters can be expressed as 800 kg m/s^2 * meters = 800 kg m^2 / s^2.

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RESPONSE --> Ok so if I were to say from now on Newton meters as a unit you would interpret that as kg m^2/s^2 ?

You need to work out the units with everything you do, and you should work them out in terms of the fundamental units kg, m and s. In this case Newtons * kg is kg m/s^2 * kg = kg^2 m / s^2, which is not kg m^2/s^2 and is not Newton meters. However this was just a careless error on your part; except for the unit error your procedure was correct. You just put down 20 kg when you meant 20 meters, most likely just a typo.

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19:53:47 `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?

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RESPONSE --> They the exact same value.

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19:54:52 The change in the quantity Fnet * `ds is 800 kg m^2 / s^2 and the change in 1/2 m v^2 is 800 kg m^2 / s^2. The quantities are therefore the same. This quantity could also be expressed as 800 Newton meters, as it was in the initial calculation of the less question. We define 1 Joule to be 1 Newton * meter, so that the quantity 800 Newton meters is equal to 800 kg m^2 / s^2 and also equal to 800 Joules.

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RESPONSE --> Ok I can see where you are going. The measure of Newton meters is equal to Joules.

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20:10:56 `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 --> Ok I won't restate the given quantities, but can solve the vf with this: vf^2 = v0^2 +2a'ds vf = sqrt (( 9 m/s )^2 + 2 ( 4 m/s^2 ) ( 20 m ) vf = sqrt ( 81 m^2/s^2 + 160 m^2/s^2 ) vf = 15.52 m/s Ok now the change in value of 1/2 m v^2: = .5 ( 10 KG )( 9 m/s )^2 = 405 Joules Now the end: = .5 ( 10 kg )( 15.52 m/s )^2 = 1204.35 Joules Now for Fnet * 'ds: = 40 Newtons * 20 m = 800 Newton meters I think I did this wrong again.

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20:12:48 The acceleration results from the same net force acting on the same mass so is still 4 m/s^2. This time the initial velocity is v0 =9 m/s, and the displacement is still `ds = 20 meters. We therefore obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt( (9 m/s)^2 + 2 * 4 m/s^2 * 20 meters) = +_`sqrt( 81 m^2 / s^2 + 160 m^2 / s^2) = +_`sqrt( 241 m^2 / s^2) = +_15.5 m/s (approx). For the same reasons as before we choose the positive velocity +15.5 m/s. The quantity 1/2 mv^2 is initially 1/2 * 10 kg * (9 m/s)^2 = 420 kg m^2 / s^2 = 420 Joules, and reaches a final value of 1/2 * 10 kg * (15.5 m/s)^2 = 1220 kg m^2 /s^2 = 1220 Joules (note that this value is obtained using the accurate value `sqrt(241) m/s rather than the approximate 15.5 m/s; if the rounded-off approximation 15.5 m/s is used, the result will differ slightly from 1220 Joules). The quantity therefore changes from 420 Joules to 1220 Joules, a change of +800 Joules. The quantity Fnet * `ds is the same as in the previous exercise, since Fnet is still 40 Newtons and `ds is still 20 meters. Thus Fnet * `ds = 800 Joules. We see that, at least for this example, the change in the quantity 1/2 m v^2 is equal to the product Fnet * `ds. We ask in the next problem if this will always be the case for any Fnet, mass m and displacement `ds. [Important note: When we find the change in the quantity 1/2 m v^2 we calculate 1/2 m v^2 for the initial velocity and then again for the final velocity and subtract in the obvious way. We do not find a change in the velocity and plug that change into 1/2 m v^2. If we had done so with this example we would have obtained about 205 Joules, much less than the 800 Joules we obtain if we correctly find the difference in 1/2 m v^2. Keep this in mind. The quantity 1/2 m v^2 is never calculated using a difference in velocities for v; it works only for actual velocities.]

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RESPONSE --> Ok I thought I did it wrong because I was looking for the last value to equal 1205 but I didn't realize that it was the difference of the beginning and the end.

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20:23:07 `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 --> Fnet = 100 Newtons, mass = 20 kg, 'ds = 50 meters and v0 = 4 m/s: a = 100 Newtons / 20 kg = 5 m/s^2 vf = sqrt (( 4 m/s)^2 + 2 ( 5 m/s^2 )( 50 m ) vf =sqrt ( 16 m^2/s^2 + 500 m^2/s^2 ) vf = 22.72 m/s = .5 ( 20 kg )( 4 m/s)^2 = 160 Joules = .5 ( 20 kg )( 22.72 m/s )^2 = 10324 Joules = 100 Newtons * 50 kg = 5000 Newton meters

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20:23:42 Following the same order of reasoning as used earlier, we see that the expression for the acceleration is a = Fnet / m. If we assume that v0 and `ds are known then once we have acceleration a we can use vf^2 = v0^2 + 2 a `ds to find vf. This is good because we want to find an expression for 1/2 m v0^2 and another for 1/2 m vf^2. First we substitute Fnet / m for a and we obtain vf^2 = v0^2 + 2 * Fnet / m * `ds. We can now determine the values of 1/2 m v^2 for v=v0 and v=vf. For v = v0 we obtain 1/2 m v0^2; this expression is expressed in terms of the four given quantities Fnet, m, `ds and v0, so we require no further change in this expression. For v = vf we see that 1/2 m v^2 = 1/2 m vf^2. However, vf is not one of the four given symbols, so we must express this as 1/2 m vf^2 = 1/2 m (v0^2 + 2 Fnet/m * `ds). Now the change in the quantity 1/2 m v^2 is change in 1/2 m v^2: 1/2 m vf^2 - 1/2 m v0^2 = 1/2 m (v0^2 + 2 Fnet / m * `ds) - 1/2 m v0^2. Using the distributive law of multiplication over addition we see that this expression is the same as change in 1/2 mv^2: 1/2 m v0^2 + 1/2 * m * 2 Fnet / m * `ds - 1/2 m v0^2, which can be rearranged to 1/2 m v0^2 - 1/2 m v0^2 + 1/2 * m * 2 Fnet / m * `ds = 1/2 * 2 * m * Fnet / m * `ds = Fnet * `ds. Thus we see that for any Fnet, m, v0 and `ds, the change in 1/2 m v^2 must be equal to Fnet * `ds.

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RESPONSE --> I must have done something wrong.

The problem suggested that you work the situation out in symbols, in order to prove that Fnet * `ds really is equal in all cases to the change in the quantity 1/2 m v^2.

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20:37:19 `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 --> First the acceleration: a = 12 Newtons / 48 kg = .25 m/s^2 vf = sqrt (( 0 m/s)^2 + 2( .25 m/s^2 )( 100 m ) vf = sqrt ( 50 m^2/s^2 vf = 7.07 m/s KE = .5 ( 48 kg )( 0 m/s )^2 KE = 24 Joules KE = .5 ( 48 kg )( 7.07 m/s )^2 KE = 1200 Joules 'dWnet = 12 Newtons * 100 m = 1200 Newton meters

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20:37:29 The work done by a 12 Newton force acting through a displacement of 100 meters is 12 Newtons * 100 meters = 1200 Newton meters = 1200 Joules. A 48 kg object subjected to a net force of 12 Newtons will accelerate at the rate a = Fnet / m = 12 Newtons / 48 kg = .25 m/s^2. Starting from rest and accelerating through a displacement of 100 meters, this object attains final velocity vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt( 0^2 + 2 * .25 m/s^2 * 100 m) = +-`sqrt(50 m^2/s^2) = 7.1 m/s (approx.). Its KE therefore goes from KE0 = 1/2 m v0^2 = 0 to KEf = 1/2 m vf^2 = 1/2 (48 kg) (7.1 m/s)^2 = 1200 kg m^2/s^2 = 1200 Joules. This is the same quantity calculated usin Fnet * `ds.Thus the change in kinetic energy is equal to the work done.

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20:41:20 `q008. How much work is done by the net force when an object of mass 200 kg is accelerated from 5 m/s^2 to 10 m/s^2? Find your answer without using the equations of motion.

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RESPONSE --> If the object accelerated from 5 m/s^2 to 10 m/s^2 then that means the 'ds would be 5 meters in 1 second 'dt. So the 'dWnet = 200 kg * 5 meters = 1000 Newton meters

kg * meters is not Newton meters. A Newton is a kg * m/s^2, so a Newton meter is a kg m^2 / s^2. This is clearly different from a kg meter.

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20:42:35 The work done by the net force is equal to the change in the KE of the object. The initial kinetic energy of the object is KE0 = 1/2 m v0^2 = 1/2 (200 kg) (5 m/s)^2 = 2500 kg m^2/s^2 = 2500 Joules. The final kinetic energy is KEf = 1/2 m vf^2 = 1/2 (200 kg)(10 m/s)^2 = 10,000 Joules. The change in the kinetic energy is therefore 10,000 Joules - 2500 Joules = 7500 Joules. The same answer would have been calculated calculating the acceleration of the object, which because of the constant mass and constant net force is uniform, the by using the equations of motion to determine the displacement of the object, the multiplying by the net force.

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RESPONSE --> Well I see that I took to much of the problem for granted without working it out.

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20:51:14 `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^2 to 10 m/s^2 while traveling 50 meters, then what net force was acting on the object?

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RESPONSE --> The beginning would be: KE = .5 ( 200 kg )( 5 m/s )^2 = 2500 Joules The end would be: KE = .5 ( 200 kg )( 10 m/s )^2 = 10000 Joules The change would be 7500 Joules and you could say: 'dWnet = Fnet * 'ds Fnet = 7500 Joules / 50 meters = 150 Newton meters

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20:51:52 The net force did 7500 Joules of work. Since the object didn't change mass and since its acceleration was constant, the net force must have been constant. So the work done was `dWnet = Fnet * `ds = 7500 Joules. Since we know that `ds is 50 meters, we can easily solve for Fnet: Fnet = `dWnet / `ds = 7500 Joules / 50 meters = 150 Newtons. [Note that this problem could have been solved using the equations of motion to find the acceleration of the object, which could then have been multiplied by the mass of the object to find the net force. The solution given here is more direct, but the solution that would have been obtain using the equations of motion would have been identical to this solution. The net force would have been found to be 300 Newtons. You can and, if time permits, probably should verify this. ]

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RESPONSE --> I stated it as Newton meters but it was just a habit. I understand the difference.

To be very clear you should state it as Newton meters, but then also as Joules. So your answer was correct but not quite complete.

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21:03:58 `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 a mobile obtain?

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RESPONSE --> First we can find the difference in work done: 'dWnet = Fnet * 'ds 'dWnet = 5000 Newtons * 80 meters = 400000 Newton meters Now we can find the beginning of the KE: KE = .5 ( 2000 KG )( 0 m/s)^2 = 1000 Joules Now knowing the difference will be 400000 we can find the final velocity with the same equation and solve for vf: 401000 Joules = .5 ( 2000 kg )( vf m/s)^2 vf = sqrt ( 401000 Joules / 1000 kg ) vf = 20.02 m/s

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21:05:03 The We know that the net force does work `dWnet = Fnet * `ds = 5000 Newtons * 80 meters = 400,000 Joules. We know that the kinetic energy of the automobile therefore changes by 400,000 Joules. Since the automobile started from rest, its original kinetic energy KE0 was 0. We conclude that its final kinetic energy KEf must have been 400,000 Joules. Since KEf = 1/2 m vf^2, this is an equation we can solve for vf in terms of m and KEf, both of which we now know. We can first multiply both sides of the equation by 2 / m to obtain 2 * KEf / m = vf^2, then we can take the square root of both sides of the equation to obtain vf = +- `sqrt(2 * KEf / m) = +- `sqrt( 2 * 400,000 Joules / (2000 kg) ) = +- `sqrt( 400 Joules / kg). At this point we had better stop and think about how to deal with the unit Joules / kg. This isn't particularly difficult if we remember that a Joule is a Newton * meter, that a Newton is a kg m/s^2, and that a Newton * meter is therefore a kg m/s^2 * m = kg m^2 / s^2. So our expression +- `sqrt(400 Joules / kg) can be written +_`sqrt(400 (kg m^2 / s^2 ) / kg) and the kg conveniently divides out to leave us +_`sqrt(400 m^2 / s^2) = +- 20 m/s. We choose +20 m/s because the force and the displacement were both positive. Thus the work done on the object by the net force results in a final velocity of +20 m/s.

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RESPONSE --> I forgot to figure in that multiplying 0 is 0. Another mathematical error.

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21:11:17 `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 --> You would have : KE = .5 ( 2000 kg )( 15 m/s )^2 KE = 225000 Joules And the difference would be: 'dWnet = 400000 Joules Then the problem for the end velocity would be: 625000 Joules = .5 ( 2000 kg )( vf m/s)^2 vf = sqrt ( 625000 Joules / 1000 kg ) vf = 25 m/s

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21:11:23 Again the work done by the net force is still 400,000 Joules, since the net force and displacement have not changed. However, in this case the initial kinetic energy is KE0 = 1/2 m v0^2 = 1/2 (200 kg) (15 m/s)^2 = 225,000 Joules. Since the 400,000 Joule change in kinetic energy is still equal to the work done by the net force, the final kinetic energy must be KEf = KE0 + `dKE = 225,000 Joules + 400,000 Joules = 625,000 Joules. Since 1/2 m vf^2 = KEf, we again have vf = +- `sqrt(2 * KEf / m) = +-`sqrt(2 * 625,000 Joules / (2000 kg) ) = +-`sqrt(2 * 625,000 kg m^2/s^2 / (2000 kg) ) = +-`sqrt(625 m^2/s^2) = 25 m/s.

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21:24:34 `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 --> First we need to find the 'dWnet: 'dWnet = 300 Newtons * 30 meters = 9000 Newton meters Then the beginning of the work will be 0 so the difference will be the KE for the end velocity: 9000 Joules = .5 ( 20 kg )( vf m/s)^2 vf = sqrt ( 9000 Joules / 10 kg ) vf = 30 m/s The work done by the applied force is 9000 Joules. The work done by the friction is : 'dWnet = 100 newtons * 30 meters = 3000 Joules So the work done by the net force is 9000 - 3000 = 6000 Joules

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21:25:41 The block experiences a force of 300 Newtons in its direction of motion and a force of 100 Newtons opposite its direction motion. It therefore experiences a net force of Fnet = 300 N - 100 N = 200 N. The work done by the net force is therefore `dWnet = 200 N * 20 m = 4000 Joules. The work done by the 300 Newton applied force is `dWapplied = 300 N * 30 m = 9000 Joules. The work done by friction is `dWfrict = -100 N * 30 m = -3000 Joules (note that the frictional forces in the direction opposite to that of the displacement). Note that the 6000 J of work done by the net force can be obtained by adding the 9000 J of work done by the applied force to the -3000 J of work done by friction. The final velocity of the object is obtained from its mass and final kinetic energy. Its initial KE is 0 (it starts from rest) so its final KE is KEf = 0 + `dKE = 0 + 6000 J = 6000 J. Its velocity is therefore vf = +- `sqrt(2 KEf / m) = `sqrt(12,000 J / (20 kg) ) = +-`sqrt( 600 (kg m^2 / s^2) / kg ) = +-`sqrt(600 m^2/s^2) = +- 24.5 m/s (approx.). We choose the positive final velocity because the displacement and the force are both in the positive direction.

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RESPONSE --> I started out wrong where I should have deducted the friction in the beginning.

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������}Z����ΰ���ב�Ȃ� assignment #001 yE�܃��������݌��� Liberal Arts Mathematics I 11-22-2005

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18:34:03 Problem Number 1 A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)? Show how you could use a graph of velocity vs. time to obtain your results. Directly reason out your results using the concept of rate. If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?

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RESPONSE --> In the 0 - 9 s time slot: v0 = 4.5 m/s, vf = 10 m/s, 'dv = 5.5 m/s, 'dt = 9s, vAve = 7.25 m/s, aAve = 0.61 m/s^2 and 'ds = 65.25 m. In the 9 - 13 s time slot: v0 = 10 m/s, vf = 2.25 m/s, 'dv = - 7.75 m/s, 'dt = 4 s, vAve = 6.125 m/s, aAve = - 1.9375 m/s^2 and 'ds = 24.5 m. You would say that the slope of the second time interval would have the greatest slope.

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18:39:39 Problem Number 2 We set up a straight ramp and measure the time required to coast down the ramp, from rest, at various slopes. The differences in elevation between one end and the other, for the different slopes, are 1.8, 4.2 and 6.7 cm. The time required for a cart to coast 78 cm down the ramp, starting from rest, is 2.722297 seconds on the first incline, 2.518101 seconds on the next, and 2.6606 seconds on the last. How well do these results support the hypothesis that acceleration is linearly dependent on slope?

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RESPONSE --> Well I don't really know how to put it into words but when I do the division of rise over run I get a value that increases size over the three problems.

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18:40:25 STUDENT RESPONSE: 1.8cm 2.722297s 4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowest The middle ramp has the fastest time down the ramp These results show that acceleration is fastest down the middle ramp (4.2cm) INSTRUCTOR COMMENT ON STUDENT RESPONSE: ** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments. You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **

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18:55:50 Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?

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RESPONSE --> We know that 'dt = 3.8 s, vf = 6.31579 cm/s, 'ds = 50 cm and vAve = 13.16 cm/s. Now we can find v0 by: 'ds = ( vf + v0 ) / 2 * 'dt 50 cm = ( 6.31579 cm/s + v0 ) / 2 * 3.8 s 100 cm / 3.8 s = 6.31579 cm/s + v0 26.3157 cm/s - 6.31579 cm/s = v0 v0 = 20 cm/s So the 'dv would be - 13.68 cm/s and the aAve would = - 13.68 cm/s / 3.8 s = - 3.601 cm/s^2.

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18:56:15 STUDENT RESPONSE: .'ds=50cm vf=6.31579cm/s 'dt = 3.8s The average velocity on the ramp is 50cm/3.8s =13.16m/s The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00 ** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation. The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have vAve = (v0 + vf) / 2.. You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify. You get v0 = 2 vAve ? vf, then substitute. This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. ** The accleration is 6.31579-20/3.8 = -3.601m/s/s If your v0 was correct this would be right **

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19:14:07 Problem Number 4 A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?

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RESPONSE --> First I will solve the free fall values. Given we have 'dt = 95 cm, aAve = 9.8 cm/s^2 and v0 = 0 cm/s. So we can use : vf^2 = v0^2 + 2a'ds vf = sqrt(( 0 cm/s)^2 + 2( 9.8 cm/s^2) ( 95 cm)) vf = sqrt ( 2 (931 cm^2/s^2)) = 1862 cm^2/s^2 vf = 43.15 cm/s So the 'dv = vf and 'dt = 43.15 cm/s / 9.8 cm/s^2 = 4.4 s.

This would be correct except that g = 980 cm/s^2, not 9.8 cm/s^2. 980 cm/s^2 = 9.8 m/s^2, since 1 m = 100 cm. If you correct the acceleration of gravity you end up with vf = 431.5 cm/s and `dt = .44 s.

The 18 cm/s is in the horizontal direction. The horizontal motion is independent of the vertical. For an ideal projectile the horizontal acceleration is 0 so horizontal velocity is constant. So in the horizontal direction the projectile moves at a constant 18 cm/s for .44 sec. How far does it therefore move in the horizontal direction?

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19:14:49 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.

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19:20:18 Problem Number 5 What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?

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RESPONSE --> We know v0 = 0 cm/s, 'ds = 40 cm and vf = 13.9 cm/s. The vAve = 6.95 cm/s , 'dv = 13.9 cm/s and 'dt = 5.76 s. So the aAve = 2.41 cm/s^2.

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19:20:42 STUDENT RESPONSE: 'ds=40cm v0=0 vf=13.9cm/s If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2 40 * a = 193.21 cm^2/s^2 / 2 a = 96.61 cm^2/s^2 / (40 cm) a = 2.415cm/s^2 ** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 ? v0^2) / 2, as you say, so a = (vf^2 ? v0^2) / (2 `ds). This is what you did (good job) but be careful to state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I wouldn?t use this equation to solve for `dt, since the equation is quadratic in `dt. At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler. Or you could just figure average velocity and divide into displacement. Either way would be fine. ** 40= +.5 2.415cm/s/s dt^2 dt = sqrt 40/ ** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt: Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **

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���F�wM�ӴD߯����꤁�Y|�� assignment #013 yE�܃��������݌��� Physics I 11-22-2005

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21:58:22 Query gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel

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RESPONSE --> The figure I came up with was way off the book's answer: Mass = 7 g, v0 = 0, vf = 125 m/s and 'ds = .8 m. So I solved for the acceleration which I had as 9765.625 m/s^2. So mine is no where near.

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21:59:15 ** The initial velocity of the bullet is zero and the final velocity is 175 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.11 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **

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RESPONSE --> Ok but the book says 68.4 N.

The solution given here is for the former edition's version of the text problem, which had 175 cm/s and .7 m rather than 125 cm/s and .8 m. Looks like I revised the statement of the problem didn't correctly revise the solution. I actually used the 125 cm/s, but I used the old .7 m rather than the new .8 m. I believe your acceleration is correct for the quantities given in the text. Had you multiplied your acceleration by .007 m I believe you would have obtained the book's 68.4 meters.

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21:59:18 univ phy *&*& now a parachuist *&*&problem 4.38 (4.34 10th edition) fish on balance, reading 60 N when accel is 2.45 m/s^2 up; find true weight, situation when balance reads 30 N, balance reading when cable breaks What is the net force on the fish when the balance reads 60 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read when the cable breaks?

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21:59:20 ** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then Fnet is still m * a and we have m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. **

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21:59:43 STUDENT QUESTION: I had trouble with the problems involving tension in lines. For example the Fish prob. Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.

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22:00:04 ** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. **

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