Download 7-1-work - High Point University

Work
Objective: To calculate work on an object by a force that acts through a certain displacement; to relate
work to energy; to understand the results of positive work and negative work.
Introduction
The momentum principle tells us that the net force times the time interval during which it acts is equal to
the change in the momentum of an object. The quantity, net force times time interval, is called impulse.
What about net force times displacement? If the net force on an object acts through a certain displacement, is the product, net force times displacement, significant? Yes, it is!
To explore this idea in more detail, let’s consider a hypothetical situation–after all, you should always
first understand a simple situation before attempting to understand a complex situation. Suppose you have
a space probe “at rest” in space, far from any objects it can interact with. An engine attached to the probe
fires, and the probe is displaced an amount ∆r in the direction its engine fired.
What will you observe? You will certainly notice the space probe speed up, and the greater the displacement through which the force acts, the greater the final speed of the probe. And what will happen when the
thruster stops firing (i.e. F~engine = 0)? The speed will be constant.
It turns out that there is single concept that unites the apparently related variables of the thrust of the
engine (and displacement through which it acts), the amount and type of fuel on board (as well as the engine
and how well it performs), and the speed of the probe. That concept is energy!
• The energy associated with the speed of the probe is called kinetic energy.
• The energy associated with the amount and type of fuel is called chemical energy.
• The performance of the thruster engine in converting chemical energy of the fuel to the kinetic energy
of the probe is related to the work done on the probe by the thruster.
Work
The mechanical energy (in this case it is kinetic energy) transferred to the probe is equal to the work done
by the engine. If the thruster exerts twice the force on the probe, then it does twice the work. Or, if it acts
through twice the displacement of the probe, it does twice the work. Therefore, work is proportional to both
force and displacement. We define it as the product of force and displacement. But how do we multiply
vectors?
Calculating Work
First, note that only force components parallel to the displacement through which the force acts can do
work.
Method 1
Work = (Component of force parallel to displacement) (Magnitude of displacement through which it acts)
(1)
If the component of force parallel to displacement is in the same direction as the displacement, the work
done is positive. This means that the work done by the force increased the mechanical energy of the probe.
If the component of the force parallel to displacement is in the opposite direction as the displacement,
the work done is negative. This means that the work done by the force decreased the mechanical energy of
the probe; perhaps the probe slowed down.
You can also write this as
~
W ork = F cosθ|∆r|
(2)
where θ is the angle between F~ and ∆r (draw the vectors tail to tail so you can figure out what the angle is
between them).
Method 2
You can also write the force as its x, y, z components and the displacement as its x, y, z components. Then,
calculate the work done by the force as it acted through the x-displacement, the work done by the force as
it acted through the y-displacement, and the work done by the force as it acted through the z-displacement.
Then, you add them up. (Remember that work is a scalar, so you are just adding numbers!)
W ork = Fx ∆x + Fy ∆y + Fz ∆z
(3)
This type of vector multiplication is called a dot product. In general, the dot product of any two vectors
is
~·B
~ = Ax Bx + Ay By + Az Bz = |A||
~ B|cos(θ)
~
A
where theta is the angle between the vectors
(4)
Note that the result is a scalar!
Non-constant forces or non-straight-line displacement
What if a force acts through a curved path? Or what if the force is not constant during the displacement it
acts through?
Then, you have to break up the path into small pieces, treating each piece as a straight-line and treating
each piece as if the force is constant through the small displacement. Find the work done by the force during
each small piece, and add up the work done during each small displacement to get the total work done by
the force – that is an integral!
If the small displacement is d~r, then the work done by the force during a small displacement is
~
dW = F~ · dr
(5)
To find the work done by the force during the entire displacement, add up each dW –in other words,
integrate.
Z
W =
f
~
F~ · dr
(6)
i
Application
1. A 0.035-kg model rocket, a long time after its engine has burned out, falls straight downward at its
terminal speed toward the earth. Suppose it falls at its terminal speed for a distance of 20.0 m.
(a) What is the work done by gravitation on the rocket during this interval?
(b) What is the work done by drag (of the air on the rocket) during this interval?
2. Suppose that when the toy rocket in the previous question is launched it is displaced vertically a distance
of 2.5 m when the engine burns out. The average thrust of the engine during that displacement is 3.0
N.
(a) What is the work done by thrust on the rocket?
(b) What is the work done by gravitation on the rocket?
3. You pull your sister 8.0 m on a sled by tugging on a rope at an angle of 35◦ with a constant force of
magnitude 50.0 N. Her mass is 10.0 kg and the mass of the sled is 10.0 kg, and the temperature outside
is 25◦ F. What is the work done by the rope on the sled/sister system?
4. A space probe’s thruster acts through the displacement < 1e5, −2e5, 9e4 > m. The force of the
thruster is < −1e2, 4e2, 2e2 > N. The force is constant and the motion of the probe is along a straight
line.
(a) What is the work done by the thruster?
(b) Why is the displacement vector not in the same direction as the force of the thruster?
5. A non-constant force < 2x2 , y, 0 > N acts through a straight-line displacement < 2, 0, 0 > m. What
is the work done by the force during the displacement?
6. If the Moon is in a circular orbit around the Earth, what is the work done on the Moon by the
gravitational force of the Earth on the Moon?