course Phys 201 miscalculated on 2 and 3 ŖPLHQassignment #013
......!!!!!!!!...................................
14:08:53 `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.
......!!!!!!!!...................................
RESPONSE --> m=10kg Fnet=40N s=20m Vo=0M/S Vf=? dW=F*ds =40N*20m dW=800J=dKE Vf=sqrt.((2(KE))/m) =sqrt.((2(800J)/(10kg)) =sqrt.(1600/10kg) =sqrt.(160m/s) Vf= 12.65m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:20:59 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:29:01 `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.
......!!!!!!!!...................................
RESPONSE --> I know that 1/2mv^2=KE but I am not sure how to get to the value with just the displacement. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:32:54 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.
......!!!!!!!!...................................
RESPONSE --> I don't understand were you got the two quanities for Vo and Vf? Please go into more detail on that. I also don't understand were you go the 10kg. self critique assessment: 1
.................................................
......!!!!!!!!...................................
15:14:12 `q003. Find the value of the quantity Fnet * `ds for the present example, and express this quantity in units of kg, meters and seconds.
......!!!!!!!!...................................
RESPONSE --> i am not sure about this question either confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:17:00 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.
......!!!!!!!!...................................
RESPONSE --> I understand this problem I was lacking the 40N and didn't understand where to get it, but I understand now. self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:21:17 `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
......!!!!!!!!...................................
RESPONSE --> Fnet*ds and the change in (1.2mv^2) is = to dW by the object+ dKE=0 confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:24:21 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:44:19 `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.
......!!!!!!!!...................................
RESPONSE --> Vf^2=Vo*2a*ds =9m/s*2(4m/s^2)(20m) Vf^2= sqrt.1440m/s vf=37.94m/s 1/2mv^2 1/2(10kg)(9m/s)^2 1/2*10kg*81m/s 1/2*810=405m/s 1/2(10kg)(37.94)=189.7 Fnet*ds 40N*20m =800J confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:59:29 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.]
......!!!!!!!!...................................
RESPONSE --> 3 self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:16:02 `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.
......!!!!!!!!...................................
RESPONSE --> I figured out the problem and it did not come out the same. I think I am doing something wrong though because when I did the previous provlem both of my values were off by 20. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:16:32 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:41:59 `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.
......!!!!!!!!...................................
RESPONSE --> dW=Fnet*ds =12N*100m dW=1200J Vf=sqrt.Vo^2+2(a)(ds) =sqrt.0^2+2(.25m/s^2)(100m) =sqrt..5m^2/s^2(100m) =sqrt.50m^2/s^2 Vf=7.07m/s a=F/m =12N/48kg a=.25m/s^2 dKE=1/2mv^2 Vo=1/2(48kg)(0)^2 =0m/s Vf=1/2(48kg)(7.07m/s) =1/2(48kg)(50m/s) Vf=1200J Vf-Vo=dV 1200m/s-0m/s=1200m/s dW=dKE=1200J confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:42:11 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:45:42 `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.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
16:49:53 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.
......!!!!!!!!...................................
RESPONSE --> i understand how to sole the problem but how do you find the acceleration given the velocities and mass? self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:01:43 `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?
......!!!!!!!!...................................
RESPONSE --> dW=Fnet*ds Fnet=dW/ds =7500J/50m Fnet=150N confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:02:26 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. ]
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:14:21 `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?
......!!!!!!!!...................................
RESPONSE --> I don't think this is right, but I will try anyways Vf=sqrt.Vo^2+2(a)(ds) =sqrt.0m/s^2+2(2.5m/s^2)(80m) =sqrt.2(2.5m/s^2)(80m) =sqrt..5(80m) =sqrt.(40m^2/s^2) Vf=6.32m/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:36:21 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.
......!!!!!!!!...................................
RESPONSE --> i understand the equation. self critique assessment:
.................................................
......!!!!!!!!...................................
18:08:38 `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?
......!!!!!!!!...................................
RESPONSE --> Vf=sqrt. Vo^2+2(a)(ds) =sqrt.15m/s+2(2.5m/s^2)(80m) =sqrt.225m/s+2(5m/s^2)(80m) =sqrt.225m^2/s^2+400m =sqrt.625m^2/s^2 =25m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:09:05 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:26:56 `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?
......!!!!!!!!...................................
RESPONSE --> W=Fnet*ds =200N*30m Wnet=6000J W=Ffrict*ds = -100N*30m Wfrict.=-3000J W=Fapp.*ds =300N*30m Wapp.=9000J confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:32:10 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 * 30 m = 6000 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.
......!!!!!!!!...................................
RESPONSE --> i forgot to answer the last part but i got the same answer. self critique assessment: 3
.................................................
藶[}z assignment #020 020. Forces (inclines, friction) Physics II 05-12-2009
......!!!!!!!!...................................
14:06:26 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
......!!!!!!!!...................................
RESPONSE --> Because the first mass is sitting on the table the gravitational force is zero. You would use the the second mass F=m2*a =2kg*9.8m/s^2 =19.6N total mass=7kg a=19.6m/s^2/7kg =2.8m/s^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:06:34 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
......!!!!!!!!...................................
RESPONSE --> i understand this question self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:16:07 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
......!!!!!!!!...................................
RESPONSE --> 49N*.10=4.9N 19.6N-4.9N=14.7N 14.7N/7kg=2.1m/s^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:16:13 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:29:19 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
......!!!!!!!!...................................
RESPONSE --> x axis=49N*cosin(282 deg) =10.19N y axis= 49N*sin(282 deg)=-47.9N 48N*.10=4.8N in the opposite direction of motion. 10.19N=-4.8N=5.39N 5.39N+19.6N= 24.99N 24.99N/7kg=3.57 m/s^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:29:25 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:14:22 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
......!!!!!!!!...................................
RESPONSE --> Vo=12m/s ds=3m a=9.8m/s^2 Vf^2=Vo^2+2a*ds Vf=sqrt. 19.6m/s^2*3m =sqrt. 58.8m^2/S^2 =-7.67m/s in the y axis direction =12m/s in the x axis direction M=sqrt. 12m/s^2+-7.7m/s^2 =144m/s+59.29m/s =Sqrt. 209.39m/s =14.2m/s tan-1(-7.7m/s/12m/s)=-32.7deg. 360 deg +-32.7deg=-327.3 deg. confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:14:34 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
......!!!!!!!!...................................
RESPONSE --> i understand this question self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:37:51 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
......!!!!!!!!...................................
RESPONSE --> Vo= 20m/s @ 30deg a vert= -9.8m/s^2 ds=-12m Vo in the y direction=20m/s*sin(30deg)=10m/s Vf=sqrt. 10m/s^2+2* (-9.8m/s^2)*-12m =sqrt. 10m/s^2+-19.6m/s^2*-12m =sqrt. 10m/s^2+235m^2/s^2 =sqrt. 245.2m/s Vf=15.6m/s in downward direction vAve=10m/s+-15.6m/s/2 =-2.8m/s dt=ds/vAve =-12m/-2.8m/s dt=4.29s confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:37:55 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:42:13 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
......!!!!!!!!...................................
RESPONSE --> dt= 4.29s Vhorz.= 20m/s*cosine(30deg) =17.3m/s 17.3m/s*4.29s= 74.2m confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:42:38 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
......!!!!!!!!...................................
RESPONSE --> I have a feeling in the prior problem a simple math error was made. self critique assessment: 3
.................................................