Nֽ݉xΟG Student Name: assignment #013
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09:04:50 `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 --> a = Fnet / m a = 40 Newtons / 10 kg a = 4 m/s^2 Use this equation to find vf:vf^2 = v0^2 + 2 a `ds
vf = `sqrt( v0^2 + 2 a `ds ) vf= `sqrt ( 0 + 2 * 4 m/s^2 * 20 meters) vf= `sqrt(160 m^2 / s^2) vf= +-12.7 m/s.................................................
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09:05:00 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 --> ok
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09:12:32 `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 it is 0. At the end it is 12.7 m/s. It changes by 12.7 m/s.
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09:14:16 02-22-2006 09:14:16 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.......!!!!!!!!...................................
NOTES -------> I think I misunderstood the last part of the question. initial value 1/2 (10 kg) (0 m/s)^2 = 0 final value 1/2 (10 kg)(12.7 m/s)^2 = 800 kg m^2 / s^2
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09:15:35 `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 = 40 Newtons * 20 meters = 800 Newton meters
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09:15:41 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 --> ok
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09:16:25 `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.
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09:16:51 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 --> ok
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09:25:54 `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 --> a still = 4 m/s^2 'ds still = 20 meters vf = `sqrt( v0^2 + 2 a `ds) vf =`sqrt( (9 m/s)^2 + 2 * 4 m/s^2 * 20 meters) vf = `sqrt( 81 m^2 / s^2 + 160 m^2 / s^2) vf = `sqrt( 241 m^2 / s^2) vf = +-15.5 m/s 1/2 mv^2 is initially 1/2 * 10 kg * (9 m/s)^2 = 420 Joules final value 1/2 * 10 kg * (15.5 m/s)^2 = 1220 Joules Total change of 800 Joules Fnet * `ds = 800 Joules
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09:25:59 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 --> ok
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09:43:48 `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 --> Do this the same way we did the other one just use different numbers.
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09:43:53 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 --> ok
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09:51:08 `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 --> 'dWnet= 1200 Newton meters = 1200 Joules a = Fnet / m a= 12 Newtons / 48 kg a= .25 m/s^2 vf = `sqrt( v0^2 + 2 a `ds) vf = `sqrt( 0^2 + 2 * .25 m/s^2 * 100 m) vf = `sqrt(50 m^2/s^2) vf = 7.1 m/s KEO starts out as 0 KEf = 1/2 m vf^2 KEF = 1/2 (48 kg) (7.1 m/s)^2 KEF= 1200 kg m^2/s^2 KEF= 1200 Joules So they are the same.
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09:51:11 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 --> ok
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09:53:08 `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 --> KE0 = 1/2 m v0^2 KEO= 1/2 (200 kg) (5 m/s)^2 KEO= 2500 kg m^2/s^2 KEO= 2500 Joules KEF = 1/2 m vf^2 KEF= 1/2 (200 kg)(10 m/s)^2 KEF= 10,000 Joule Change in energy = 10,000 Joules - 2500 Joules = 7500 Joules
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09:53:12 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 --> ok
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09:54:56 `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 --> Fnet = `dWnet / `ds = 7500 Joules / 50 meters = 150 Newtons
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09:55:00 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 --> ok
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09:58:57 `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 --> dWnet = Fnet * `ds dWnet= 5000 Newtons * 80 meters dWnet = 400,000 Joules Use this equation to solve for vf KEf = 1/2 m vf^2 vf = `sqrt(2 * KEf / m) vf = `sqrt( 2 * 400,000 Joules / (2000 kg) ) vf= `sqrt( 400 Joules / kg) vf = 20 m/s
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10:01:54 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 --> Wdnet = 400,000 Joules KE0 = 1/2 m v0^2 KE0= 1/2 (200 kg) (15 m/s)^2 KE0= 225,000 Joules KEf = KE0 + `dKE KEf= 225,000 Joules + 400,000 Joules KEf= 625,000 Joules vf = `sqrt(2 * KEf / m) vf = `sqrt(2 * 625,000 Joules / (2000 kg) ) vf = `sqrt(2 * 625,000 kg m^2/s^2 / (2000 kg) ) vf = `sqrt(625 m^2/s^2) = 25 m/s
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10:08:15 `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 --> I am confused on which question I should be answering. I had to run an errand and now I can't figure out which one I am answering. I think I have already answered number 11 but it appears that is what I should be doing. Did I skip one?
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10:08:29 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 --> ok
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10:14:32 `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 --> Fnet = 300 N - 100 N = 200 N `dWnet = 200 N * 30 m = 6000 Joules `dWapplied = 300 N * 30 m = 9000 Joules `dWfrict = -100 N * 30 m = -3000 Joules KEf = 0 + `dKE = 0 + 6000 J = 6000 J vf = `sqrt(2 KEf / m) vf = `sqrt(12,000 J / (20 kg) ) vf= `sqrt( 600 (kg m^2 / s^2) / kg ) vf= `sqrt(600 m^2/s^2) vf= 24.5 m/s
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10:14:36 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 --> ok
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lw鞯R assignment #013 [뺗܃ܝ Physics I 02-22-2006
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16:08:41 prin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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RESPONSE --> Use a*m = Fnet and solve for m. m = Fnet / a 256 N / 2.30 m/s^2 = 115 kg
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16:08:51 A force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m.
Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg.......!!!!!!!!...................................
RESPONSE --> ok
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16:13:23 prin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> average velocity = (0 + 125 m/s) / 2 = 62.5 m/s time = .7 m / (62.5 m/s) = .011 s Assuming uniform acceleration a = = (125 m/s - 0 m/s) / (.11 sec) a = 11000 m/s^2 F = m a F = .007 kg * 11000 m/s^2 F = 77 N
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16:13:55 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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16:14:10 gen phy 4.08. breaking strength 22 N, accel 2.5 m/s^2 breaks line. What can we say about the mass of the fish?
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RESPONSE --> N/A
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16:14:16 The fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - m g, where m is the mass of the fish.
To accelerate a fish of mass m upward at 2.5 m/s^2 the net force must be Fnet = m a = m * 2.5 m/s^2. Combined with the preceding we have the condition m * 2.5 m/s^2 = T - m g so that to provide this force we require T = m * 2.5 m/s^2 + m g = m * 2.5 m/s^2 + m * 9.8 m/s^2 = m * 12.3 m/s^2. We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus m * 12.3 m/s^2 > 22 N. Solving this inequality for m we get m > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg. The fish has a mass exceeding 1.8 kg.......!!!!!!!!...................................
RESPONSE --> N/A
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16:14:24 univ phy 4.38*parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist?
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RESPONSE --> N/A
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16:14:31 Describe the free body diagram you drew.
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RESPONSE --> N/A
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16:14:36 The weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is
-540 + 620 N = 80 N. Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx........!!!!!!!!...................................
RESPONSE --> N/A
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16:14:42 univ phy (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?......!!!!!!!!...................................
RESPONSE --> N/A
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16:14:46 ** 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. **......!!!!!!!!...................................
RESPONSE --> N/A
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16:14:54 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.......!!!!!!!!...................................
RESPONSE --> N/A
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16:14:59 ** 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. **......!!!!!!!!...................................
RESPONSE --> N/A
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