course phy 231
A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Why does this analysis stop at the instant of impact with the floor?
the vo is 20cm/s downward so vf= 0 so dv is 20cm/s. a is -980cm/s^2 ds=120cm.
vf=0 vave = 10/cm/s
You know v0, a and `ds for the vertical motion. The quantities you are given are not compatible with a final velocity of 0.
Of course the ball probably stops when it reaches the floor, but the part of its motion which is characterized by uniform acceleration ends at first contact with the floor, and you don't have any information about the conditions after that instant.
dv/a=dt =20cm/s/980cm/s^2=.02s a=0 in horizontal v0= 80cm/s
20 cm/s is the initial downward vertical velocity, not the change in vertical velocity.
ds=80cm/s*(.02s)=1.6cm vf= 80cm/s vave =80cm/s
the ball should have upward acceleration after hitting the floor. this will change its direction to a bounce. the effect of the bounce is not calculable at this point in our class
You need to revise your work on this question, according to my notes.
Q a 13
013. Energy
Question: **** `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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Your solution:
f/m=a =4m/s^2 vo=0 vf^2=0^2+2(4m/s^2)(20m)=root160=12.7m/s
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
i got it!
Self-critique Assessment:
Question: **** `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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Your solution:
vo to vave is 0-6.35m/s so the ave for this interval is about 3m/s.
.5(10)(3M/S)^2=45 joules
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
i thought i was supposed to evaluate the beginning of the 20 meters
Self-critique Assessment:
Question: **** `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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Your solution: 800kgm^2/s^2(20)=1600kgm^3/s^2
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary): i knew something was wrong wheni saw m^3
Self-critique Assessment:
Question: **** `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
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Your solution:
45 keand 40 fnet doesnt make sense that the ke is greater than fnet. something is missing
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary): missed acouple of details on that one. I used the 45 from my previous incorrect answer and missed the ds part.
Self-critique Assessment:
Question: **** `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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Your solution:
a=4m/s^2, vf^2=9m/s^2+2(4m/s^2)(20m)=sqrt81m/s+160m/s^2=15.5m/s.
.5(10)(15.5^2)=77.5joules
fnet ds=40m/s*20m=800joules
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
didnt square my 15.5 and didnt subract the begining and ending energies to get the displeced energy. this is important from any interval that isnt from rest
Self-critique Assessment:
Question: **** `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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Your solution:
a= m/fnet vf=sqrt(v0^2+ 2a'ds),V0=sqrt(vf^2-2a'ds)
.5m(v0)^2 - .5m(vf^2)
this will equal fnet ds if energy is channeled efficiently
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
ok i got it.
Self-critique Assessment:
Question: **** `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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Your solution:
fnet 'ds= 12n*100=1200
a=f/m=12N/48kg=.25m/s^2
Vf^2=0+2(.25m/s^2)(100m)=sqrt50=7.1m/s
ke=.5(48kg)(0)^2=0 ke= .5(48kg)(7.1m/s)^2=1200 joules
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
should be writing fnet in calculations
Self-critique Assessment:
Question: **** `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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Your solution:
ke=.5(200kg)5^2=2500joules
ke2=.5(200kg)(10m/s)^2=10,000joules
dke=7,500joules
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
easier for this problem to use our ke equation
Self-critique Assessment:
Question: **** `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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Your solution:
dw/'ds=fnet=7500Nm/50m=150N
Confidence Assessment:
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Given Solution:
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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Self-critique (if necessary):
i didnt do it but i thought about the calculations
Self-critique Assessment:
Question: **** `q010. Solve the following without using any of the equations of motion.
net force of 5,000 Newtons acts on an automobile of mass 2,000 kg, initially at rest
need to find ds or vf but cannot think of how to do this with the work equations
Query 13
013. `query 13
Question: **** `qprin 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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Your solution:
Confidence Assessment:
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Given Solution:
`aA 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.
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Self-critique (if necessary):
Self-critique Assessment
Question: **** `qprin 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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Your solution:
Confidence Assessment:
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Given Solution:
`a** 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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Self-critique (if necessary):
Self-critique Assessment
Question: **** `qgen 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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Your solution:
Confidence Assessment:
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Given Solution:
`aThe 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.
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Self-critique (if necessary):
Self-critique Assessment
Question: **** `quniv phy 4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist?
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Self-critique (if necessary):
w=mg=55kg*9.8m/s^2)=539N
a with resistance is, 620N/55kg=11.2m/s^2 and gravity is -9.8m/s^2, so total a is 1.47m/s^2 upward
Self-critique Assessment
Question: **** `qDescribe the free body diagram you drew.
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Your solution:
there is an arrow downward of 539N and a component upward of 620N . it is accelerating upward, this doesnt match what i would think should be going on.
Confidence Assessment:
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Given Solution:
`aThe 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..
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Self-critique (if necessary):
o thought i did it wrong but it was right. i guess it makes sense that slowing down would go hand in hand with upward acceleration in this case.
Self-critique Assessment
Question: **** `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction.
What is the net force on the fish when the balance reads 50 N?
What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks?
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Your solution:
the net force is said to be measured to be 50 N
the weight is from m=f/m =50n/2.45m/2^2=20.41kg
weight is then 20.41*9.8m/s^2=200N
Confidence Assessment:
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Given Solution:
`a** 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
F_net = 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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Self-critique (if necessary):
net force is 50 n/ a +g , comes from ma=50N-mg not just fift. this makes sense as i think about it more. nforce= ma gives 50/a when a is 2,45 plus the component of gravity. so the mass is 4kg.
for the 30N we know m so we plug in to find a.
no force when nothing is attatched.
Self-critique Assessment
Question: **** `qSTUDENT 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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Your solution:
the f up should be added to gravity as an absolute value not subracted. should be 22N/(4.5+9.8)=1.54kg
Confidence Assessment:
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Given Solution:
`a** 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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Self-critique (if necessary):
yeah
Self-critique Assessment
&#Good work. See my notes and let me know if you have questions. &#