Expert answer:Physics quiz
Expert answer:In class Quiz assignment It is going to be about:Chapter 6 Work and Kinetic Energy Chapter 7 Potential Energy and Energy Conservation
Chapter 8 Momentum, Impulse, and Collisions Chapter 9 Rotation of Rigid Bodiesthese are just to look over
chapter_6.pdf
chapter_7_prob_70.pdf
chapter_8_hmwk_solutions.pdf
chapter_9_prob_86.pdf
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Chapter 6
Work and Kinetic Energy
Up until now, we have assumed that the force is constant and thus, the acceleration is constant. Is there a simple technique for dealing with non-constant forces?
Fortunately, the answer is, “Yes.” In this chapter, we will introduce the following
concepts:
1. work
2. kinetic energy
3. the connection between work and kinetic energy.
1
Work
The work performed by a force in the direction of a straight-line displacement
is
W = Fs
(Constant force in the direction of straight-line displacement)
where F is the force [N] and s is the displacement [m].
The SI units of work is N · m also called a joule.
1 N ·m=1 J
1 joule
1 J = 0.7376 ft · lb
W = F s cos φ
W = F~ · ~s
Wtot
(constant force, straight line displacement)
(constant force, straight-line displacement)
X
~
~
~
=
F · ~s = F1 + F2 + · · · · ~s
1
Figure 1: Figure 6.4 from University Physics
Ex. 3 A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a
level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25. a) What
magnitude of force must the worker apply? b) How much work is done
on the crate by this force? c) How much work is done on the crate by
friction? d) How much work is done by the normal force? By gravity?
e) What is the total work done on the crate?
2
Work and Kinetic Energy
Constant force in the same direction as the displacement:
v 2 − vo2
W = F · s = ma · s = m
2s
where
K =
1 2
mv
2
s =
1 2 1 2
mv − mvo
2
2
(1)
(K = the kinetic energy)
So, the relationship between work and the kinetic energy is:
Wtot = Kf − Ki = ∆K
(The Work-Energy Theorem)
2
(2)
Ex. 24 You throw a 3.00-N rock vertically into the air from ground level. You
observe that when it is 15.0 m above the ground, it is traveling at
25.0 m/s upward. Use the work-energy theorem to find a) its speed
just as it left the ground and b) its maximum height.
3
Work and Energy with Varying Forces
Work resulting from the application of a varying force (i.e., a force that varies as a
function of x), is the sum of the “incremental works” :
Z x
W = F1 ∆x1 + F2 ∆x2 + F3 ∆x3 + · · · =
Fx dx
xo
For a constant force, W = Fx (x − xo ).
Using integral calculus, we can calculate the work done by a variable force if we
know the force as a function of x (Fx (x)).
Z
x
W =
Fx (x) dx
xo
3.1
Hooke’s Law and the work done by a spring
Hooke’s Law is applicable to a wide number of physical systems. In many systems,
the restoring force is proportional to the displacement from equilibrium. The tendency for a system to return to its “equilibrium” position” can be written as
F = −k x
Hooke’s Law
(3)
where x is the displacement from equilibrium (x = 0), and k is the spring constant.
N.B. In this section, the book correctly states that the work done on the
spring is W = 21 kx22 − 12 kx21 . This is true if you use F = kx. However,
in the application of Hooke’s Law (Eq. 3), we are more interested in
the work done by the spring on a system. This leads to the following
equation:
3
Figure 2: Figure 6.18 from University Physics showing the work done on the spring.
Zx2
W =
x1
Zx2
1
1
Fx dx = (−kx) dx = kx21 − kx22
2
2
Work done by the spring
x1
and this is consistent with Eq. 7.10 used in the next chapter.
Ex. 34 To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work
must be done. (a) What is the force constant of this spring? (b) What
magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this
spring 4.00 cm from its unstretched length, and what force is needed
to compress it this distance?
Ex. 36 A child applies a force F~ parallel to the x-axis to a 10.0-kg sled moving
on the frozen surface of a small pond. As the child controls the speed
of the sled, the x-component of the force she applies varies with the
x-coordinate of the sled as shown in Fig. E6.36. Calculate the work
done by the force F~ when the sled moves a) from x=0 to x=8.0 m;
b) from x=8.0 m to x=12.0 m; c) from x=0 to 12.0 m.
3.2
The work-energy theorem for a varying force
Finally, we take another look at the work-energy theorem for forces that vary with
position.
4
Figure 3: Figure E6.36 from University Physics 14th edition.
Zx2
W =
Zx2
F (x) dx =
x1
Zx2
max dx =
x1
m
dv
dx =
dt
x1
Zx2
m
dv dx
dx =
dx dt
x1
Zx2
m
dv
v dx (4)
dx
x1
Zv2
W =
mv dv =
1 2 1 2
mv − mv (5)
2 2 2 1
v1
So, the work-energy theorem is the same as before even though the force is not
constant. Revisit exercise 36 and calculate the final velocity of the sled, assuming
it starts from rest.
4
Power
The definition of work makes no reference to the passage of time. We define a
physical quantity called “power” that describes the rate at which work is done on
the system. Similar to work and energy, power is also a scalar quantity.
When a quantity of work ∆W is done during a time interval ∆t, the average work
done per unit time or average power is defined to be
Pav =
∆W
∆t
(average power)
5
The rate at which work is done may not be constant, in which case, we can define
the instantaneous power using the definition of the derivative:
dW
∆W
=
∆t→0 ∆t
dt
(instantaneous power)
P = lim
Ex. 56 When its 75-kW (100-hp) engine is generating full power, a small singleengine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s
(150 m/min, or 500 ft/min). What fraction of the engine power is being
used to make the airplane climb? (The remainder is used to overcome
the effects of air resistance and of inefficiencies in the propeller and
engine.)
Ex. 58 An elevator has mass 600 kg, not including passengers. The elevator
is designed to ascend, at constant speed, a vertical distance of 20.0 m
(five floors) in 16.0 s, and it is driven by a motor that can provide up
to 40 hp to the elevator. What is the maximum number of passengers
that can ride in the elevator? Assume that an average passenger has
mass 65.0 kg.
Prob. 71 A small block with a mass of 0.0600 kg is attached to a cord passing
through a hole in a frictionless, horizontal surface (Fig. P6.71). The
block is originally revolving at a distance of 0.40 m from the hole with
a speed of 0.70 m/s The cord is then pulled from below, shortening
the radius of the circle in which the block revolves to 0.10 m. At
this new distance the speed of the block is observed to be 2.80 m/s.
a) What is the tension in the cord in the original situation when the
Block has speed v = 0.70 m/s? b) What is the tension in the cord in
the final situation when the block has speed v = 2.80 m/s? c) How
much work was done by the person who pulled on the cord?
6
Prob. 72 A proton with mass 1.67 × 10−27 kg is propelled at an initial speed of
3.00 × 105 m/s directly toward a uranium nucleus 5.00 m away. The
proton is repelled by the uranium nucleus with a force of magnitude
F = α/x2 , where x is the separation between the two objects and
α = 2.12 × 10−26 N·m2 . Assume that the uranium nucleus remains
at rest. a) What is the speed of the proton when it is 8.00 × 10−10 m
from the uranium nucleus? b) As the proton approaches the uranium
nucleus, the repulsive force slows down the proton until it comes
momentarily to rest, after which the proton moves away from the
uranium nucleus. How close to the uranium nucleus does the proton
get? c) What is the speed of the proton when it is again 5.00 m away
from the uranium nucleus?
N.B. Problem 72 can be done without integration once we introduce the concept of conservative forces and potential energy functions in chapter 7.
7
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