course phy201
3/8/2010 1:41pm
Problem Set 3Set 3 Problem number 1
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object is moving too fast, and is trying to slow down.
It is pushing against another object in front of it, which is acting to slow it down.
It pushes with a force of 4 Newtons.
How much work does it do on the other object over a distance of 19 meters?
How much energy of motion does it therefore lose to the other object?
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'dw=fave * 'ds
ds= 19
f=4
'dw= 76
Solution
Work is the product of parallel force and displacement. In this case the force is directed against an object in front of the first object, and must therefore be in the direction of motion. The force is therefore parallel to the displacement.
The force is 4 Newtons and displacement is 19 meters.
The work is therefore the product ( 4 Newtons)( 19 meters) = 76 Joules.
Generalized Solution
The precise definition of the work done by a force is the product of the force component parallel to motion and the displacement through which the force acts.
In the present problem, force and displacement are assumed to be parallel.
In this case we can say that `dW = F `ds.
We say that in order to perform this work `dW, we must expend energy `dW:
The work done and the energy required to do the work are equivalent.
Usually in the process of doing work `dW, the process is not 100% efficient and energy in excess of `dW is required. In no case can the energy required by less than `dW.
Explanation in terms of Figure(s), Extension
The figure below depicts a force F acting on a moving object along a straight-line path from point A to point B.
The force is considered uniform in the direction from A to B.
The work done by the force is `dW = F `ds.
Figure(s)
Set 3 Problem number 2
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object must gain 16 Joules of energy from a force of 6 Newtons being exerted on it parallel to its direction of motion.
If the object does not dissipate any of the energy added to it, then over what distance must it be pushed?
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'dw=16j
f=6newtons
'dw=f 'ds
16= 6* 'ds
ds= 16/6= 2.67
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Solution
It is assumed that there is no dissipation of energy, so the work done must be equal to the 16 Joules of energy gained from the applied force.
Since work equals parallel force times displacement, 16 Joules is equal to 6 Newtons multiplied by the distance.
This could be written as an equation:
16 J = ( 6 N)(dist).
Solving the equation we find that
dist = 16 J/( 6 N) = 2.666667 meters.
Generalized Solution
The precise definition of the work done by a force is the product of the force component parallel to motion and the displacment through which the force acts.
In the present problem, force and displacement are assumed to be parallel and in the same direction.
In this case, since the work done is equal to the energy expended we can say that `dW = F `ds.
Solving we see that F = `ds / `dW.
Explanation in terms of Figure(s), Extension
The figure below depicts a force F acting on a moving object along a straight-line path from point A to point B.
The force is considered uniform in the direction from A to B.
The work done by the force is `dW = F `ds.
Figure(s)
Set 3 Problem number 3
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
How much force, exerted parallel to the direction of motion, is necessary when pushing an object 6 meters in order to use up 90 Joules of energy from the pushing source?
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'ds=6meters
'dw=90joules
'dw= f 'ds
90joules = f '6meters
f= 15newtons
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Solution
The work done in the 6 meter distance must be equal to the 90 Joules of energy used up.
Since work equals parallel force times displacement, 90 Joules is equal to 6 meters multiplied by the force.
This could be written as an equation
90 J = ( 6 m)(force).
Solving we find that
force = 90 J/( 6 m) = 15 Newtons.
Generalized Solution
Since work is the product of parallel force and distance, the work necessary to dissipate energy `dE over a distance `ds by a parallel force F will be
`dW = E = F `ds and
F will therefore be the quotient
F = `dW / `ds, or equivalently F = `dE / `ds.
Explanation in terms of Figure(s), Extension
The figure below shows the relationships among force, displacement and work for the case of force parallel to displacement.
The relationship F = `dW / `ds tells us, for example, that if we wish to do a lot of work `dW over a long distance `ds we must exert only a little force.
The relationship `ds = `dW / F tells us, for example, that if we wish to do of work `dW using a relatively large force we need only exert that force through a relatively short distance.
The relationship `dW = F `ds tells us, for example, that the more force we exert and the greater the distance over which we exert it, the more work we do.
Figure(s)
Set 3 Problem number 5
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
If no energy is dissipated (e.g., against friction or air resistance) when you push an object, the increase in the energy of motion (kinetic energy, abbreviated KE) of the object is equal to the work you do on it. Since there is always some dissipation of energy in the real world, this is an ideal case.
If you exert a force of 540 Newtons while pushing an object 40 meters, with no energy dissipated, what will be its increase in kinetic energy?
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f=540newtons
'ds=40 meters
'dw= f'ds
'dw= 540newtons * 40 meters
'dw= 21600joules
(ok, since no energy was dissipated the work is equal to the increase in kinetic energy)
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Solution
The work done on the object will be the product of force and distance, or ( 540 Newtons)( 40 meters) = 21600 Joules.
Since no energy is dissipated, this is equal to the increase in kinetic energy.
Generalized Solution
The idea here is that the work done on the object is equal to its change `dKE in kinetic energy.
Since the object exerts a force equal and opposite to the force exerted on it, which is the force doing the work to change the object's KE, the object must do work equal and opposite to the work done on it.
When the force exerted on the object is in the direction of the object's motion, it tends to do positive work on the object and thus to speed the object up.
In this case the force exerted by the object is opposite to the direction of motion and we say that the object does negative work.
So when the object does negative work, work is actually done on it and it speeds up.
On the other hand if the force exerted on the object is directed opposite to the object's direction of motion it will tend to slow the object down.
In this case the force exerted by the object is in the direction of motion and the object does positive work, expending kinetic energy in the process (the energy to do the work has to come from somewhere; in this case it comes from kinetic energy).
We say in general that if `dW is the work done by the object (equal and opposite to the work done by the net force acting on the object), then
`dW + `dKE = 0.
This equation ensures that, for example, when the object does positive work the kinetic energy change must be negative (i.e., the kinetic energy must decrease).
Explanation in terms of Figure(s), Extension
Figure description: The figure below depicts in bar-graph form the relationship between the kinetic energy of an object and the work done by the object.
The first blue bar depicts the kinetic energy of the object before it does positive work, while the second depicts the kinetic energy after the work is done.
The red portion of the second bar depicts the work done by the object.
The total heights of the two bars are identical, showing that the decrease in kinetic energy is equal to the work done by the object.
Note that when the object does positive work, the force acting on the object does negative work, thereby slowing the object and decreasing its kinetic energy.
This figure depicts the situation when the work done by the object is positive.
A similar picture for the case where the object does negative work would be harder to construct and to understand, but it is not difficult to understand that when the object does negative work the work done on it is positive, and positive work done on an object increases its kinetic energy.
Figure(s)
Set 3 Problem number 6
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Problem
Solution
Generalized Solution
Problem
If you push an object 50 meters, using a force of 40 Newtons, then what will be its increase in kinetic energy if in the process it dissipates 1400 Joules to friction (assume that the only forces acting are your push and friction)?
'ds=50 meters
f= 40 newtons
'dw= 1400 joules
ok, I had to look at solution for this one, but we must find the 'dw for the the 'ds of 50 meters first
'dw= 40 newtons * 50 meters
'dw = 2000joules
then subtract to find difference
2000-1400joules= 600joules
The difference will find the increase in energy from 1400 joules to 2000joules
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Solution
The work done is ( 50 meters)( 40 Newtons) = 2000 Joules.
After dissipating 1400 Joules, there are 600 Joules left.
This is the increase in kinetic energy.
Generalized Solution
The increase in KE is equal to the work `dW done on the object by the 'pushing' or 'pulling' force, less that dissipated by other forces.
In this case, friction dissipates energy `dEfrict.
The KE increase is therefore `dW - `dEfrict.
The pushing force is in the direction of motion, the frictional force opposite to the direction of motion.
The object's equal and opposite response to the pushing force is opposite to its direction of motion so the object does negative work with this force, thereby tending to increase its KE.
The object must also exert a force opposite to the frictional force, and therefore in its direction of motion. So the object does positive work with this force, thereby tending to decrease its KE.
Set 3 Problem number 7
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Problem
Solution
Generalized Solution
Explanation in terms of Figure(s), Extension
Figure(s)
Problem
An object, initially at rest, is acted upon by a net force of 42 Newtons. The object has mass 20 kilograms.
What will be its acceleration, and what will be its speed after the first 4 seconds of acceleration?
F=m*a
f= 42 newtons
m= 20kilos
a= f/m
a= 42newtons/ 20 kilos
a=2.1 newtons/kilo
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Solution
The first question is asking for the rate at which velocity increases, which is a=F/m=( 42 Newtons)/( 20 Kg)= 2.1 meters per second per second.
Changing by 2.1 meters per second every second, in 4 seconds the increase will be ( 4 sec)( 2.1 m/sec/sec) = 8.4 m/s.
Since the object started from rest, this will also be its velocity after 4 seconds.
Generalized Solution
The rate at which velocity changes, or the acceleration, is a = F / m.
During a time interval `dt, the velocity will therefore change by a `dv = (F / m) `dt.
If the object starts from rest its final velocity will therefore be
final velocity from rest = vf = 0 + `dv = (F / m) `dt.
Explanation in terms of Figure(s), Extension
The first figure below depicts a 'flow' triangle for an object of mass m subjected to a force F.
The relationship a = F / m can be understood as saying that greater force implies proportionally greater acceleration for a given mass while greater mass implies proportionally less acceleration for a given force.
The relationship F = m * a can be understood as saying that
to achieve a given acceleration a greater force must be exerted on a greater mass and that
for a given mass a greater acceleration will require a greater force.
The relationship m = F / a tells us that for a given observed acceleration the greater the force the greater the mass being accelerated, and for a given applied force a greater acceleration implies that less mass is being accelerated.
The second figure depicts a 'flow' diagram for an object of mass m, initially with velocity v0, subjected to a force F for time interval `dt.
From the force and the mass we deduce the acceleration (the 'blue' triangle).
From the acceleration and the time interval we deduce the change `dv in velocity (the 'green' triangle).
From the change in velocity and initial velocity we obtain the final velocity (the 'purple' triangle).
Figure(s)
The only issue I had with these problems was the units......I guess its just something you have to remember??
You need to use units with every calculation, and think about what they mean and what they tell you about the situation.
This will help you become familiar with units and will increase your understanding of these various situations.