If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
013. Energy
Question: `q001. An object of mass 10 kg is subjected to a
net force of 40
Your solution:
Confidence rating:
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
a = Fnet / m = 40
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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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.
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
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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.
Your solution:
Confidence rating:
Given Solution:
Fnet = 40
Recall that a
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Question: `q004. How does the quantity Fnet * `ds and the change in (1/2 m v^2) compare?
Your solution:
Confidence rating:
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
We define 1 Joule to be 1
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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.
Your solution:
Confidence rating:
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
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.]
STUDENT COMMENT:
I rounded to 15.5m/s instead of using
sqrt(241). Will use the more
accurate value in the future.
INSTRUCTOR RESPONSE
As long as you round to the appropriate
number of significant figures, you're OK.
However in situations where you're going to be squaring the result anyway, you
introduce less roundoff error if you leave it in the radical form. In these
cases radical form is simply more convenient.
Self-critique (if necessary):
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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.
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
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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
Your solution:
Confidence rating:
Given Solution:
The work done by a 12
A 48 kg object subjected to a net force of 12
a = Fnet / m =
12
.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.
Self-critique (if necessary):
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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.
Your solution:
Confidence rating:
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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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?
Your solution:
Confidence rating:
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
[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
STUDENT COMMENT: i dont understand
why we couldnt use accel. and mass.
INSTRUCTOR RESPONSE You need to learn how to solve problems based on energy,
simply because acceleration isn't always constant and you don't always have the
information required to find acceleration. This is why you are asked to solve
the problem using energy considerations.
STUDENT COMMENT: i understand how to
sole the problem but how do you find the acceleration given the velocities and
mass?
INSTRUCTOR RESPONSE: The acceleration is not necessary to solve this
problem. It can be solved using energy considerations only.
If you were to assume a time interval, then you could find the acceleration and
the net force. Different time intervals would give you different accelerations
and different net forces, as well as different displacements; however the work
done would be the same in every case. The work done by the net force depends
only on the mass and the initial and final velocities.
Self-critique (if necessary):
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Question: `q010. Solve the following without using any of the equations of motion.
A net force of 5,000
Your solution:
Confidence rating:
Given Solution:
The We know that the net force does work `dWnet = Fnet * `ds
= 5000
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
that
a
a
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.
Self-critique (if necessary):
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Question: `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?
Your solution:
Confidence rating:
Given Solution:
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 (2000 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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Question: `q012. Solve without using the equations of motion:
A force of 300
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?
Your solution:
Confidence rating:
Given Solution:
The block experiences a force of 300
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
`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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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
Question:
`q013.
What is the velocity of a 5 kg mass whose kinetic energy is 50 Joules?
Your solution:
Confidence rating:
Question:
`q014.
A 1500 kg automobile is moving at 10 m/s.
If its kinetic
energy increases by 100 000 Joules, how fast will it be moving?
Suppose the
automobile accelerated uniformly as it gained the 100 000 Joules of kinetic
energy, which it gained in 10 seconds.
What can be determined from this information?
Indicate all the answers you can to this question, and how these
quantities can be determined.
(University
Physics students): If the energy is
added at a constant rate of 10 000 Joules / second, will the distance traveled
by the automobile be the same as, greater than or less than the distance
calculated previously.
Consider the
fact that a
constant force will not add energy at a constant rate to an object whose
velocity is changing.
Challenging question: If the distance is different how far will the automobile travel during that time?
Your solution:
Self-critique (if necessary):
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