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Transcript
```Friction, Work and the Conservation of Energy©98
Experiment 6
Objective: To demonstrate the equilibrium of forces acting on a uniformly moving object
when one of the forces is due to friction, and to show how the conservation of energy
applies to this situation.
DISCUSSION:
The block, sliding along the inclined plane shown in Fig. 1, is acted upon by three
forces. The first of these is the force of gravity, W, and is called the weight of the block.
The second is the force P of the plane on the block, and the third is the force T of the
string on the block. If the block moves at a constant velocity up or down the plane, that
is, it is not accelerated; the net force on the block is zero. Therefore, the vector sum of
these three forces is zero. We write the vector Equation for the equilibrium of forces as
W  P T  0
(1)
N
P
T
T
P
f
n
Wsi

Block

Weight
W
Figure 1. The inclined plane.

Wcos
W
Figure 2. Force components.
Breaking these forces into components parallel and perpendicular to the plane, as
in Fig. 2, we have
 W sin θ  T  f  0
(2a)
and
(2b)
Wcosθ  N  0
where  is the angle of inclination of the plane to the horizontal, f is the component of P
parallel to the plane and is called friction, and N is the component of P perpendicular to
the plane and is called the normal force. The sign convention is that components directed
up the plane are positive and those directed down the plane are negative.
6-1
For low speeds, the magnitude of the force of friction between the plane and the
block is constant and independent of the direction of motion. We can therefore use Eq.
(2a) to find f without knowing Wsin by considering uniform motion both up and down
the plane. That is, if we pull on the string with a force Tu to move the block up the plane
at a constant speed, we write Eq. (2a) as
(3a)
W sin θ  T  f
u
and if we pull on the string with a force Td which is weak enough to allow the block to
slide down the plane under the influence of its weight and at a constant speed, we write
W sin θ  T  f
(3b)
d
Subtracting Eq. (3b) from (3a) and solving for f, we have
f 
Tu  Td 
(4a)
2
On the other hand, we can add Eqs. (3a) and (3b) to eliminate friction and find the
value of Wsin  in terms of Tu and Td. We have
W sin θ 
Td  Tu 
(4b)
2
The tension Tu and Td are produced by weights hanging from the string, as shown
in Fig. 1. The weight needed to cause the block to move up the plane at a constant speed
is also the tension Tu in the string, and the weight needed to allow the block to move
down the claim at constant speed is the tension Td.
Another way of looking at the
s
motion of this system is to view it
from the standpoint of energy. As the
block is moved along the plane at a
h
s
constant speed, its potential energy
changes while its kinetic energy stays

constant. This change in potential
energy is partially matched by a
Figure 3. The conservation of energy.
change in the potential energy of the
weight hanging from the pulley. For example, if the block moves up the plane and the
weight drops, the increase in the potential energy of the block is less than the decrease in
potential energy of the weight. Since the kinetic energy of the system is constant, this net
loss in potential energy must be compensated for by some mechanism other than changes
in potential and kinetic energy. This mechanism is the energy lost through friction. To
see this relationship specifically, we refer to Fig. 3. Assume that the object moves a
distance s up along the incline. Then its increase in potential energy is mgh, where h is
the corresponding vertical height through which the block moves, and m is its mass. On
the other hand, the loss in potential energy of the weight is Tus. The net loss in potential
energy in the system is thus Tus - mgh. This loss must be equal to the work, |fs|, done by
the friction f. That is,
(5)
T s  mgh  fs
}
u
6-2
A similar equation applies if the object slides down the plane except that the loss
in potential energy of the block is greater than the gain in the potential energy of the
weight.
EXERCISES:
1.
a. Incline the plane at an angle large enough so that the unattached block
accelerates down the plane. Determine sin from appropriate trigonometric measures.
b. Measure the mass of the block.
c. Determine the force of friction acting on the sliding block by:
1) Finding the weight Tu which causes the block to move at a constant
speed up the plane.
2) Finding the weight Td which allows the block to move at a constant
speed down the plane.
3) Applying Eq. (4a).
2.
a. Find the value of Wsin  by
1) Applying Eq. (4b).
2) Using the data from Exercises (1a) and (1b).
b. Determine the percent deviation of part (2a-1) from part (2a-2).
3.
a. Measure the height h through which the block rises vertically when the weight
falls a distance s. (Note that s is also the distance the block moves along the plane.)
b. Calculate the loss in the potential energy of the weight when it falls the
distance s, the gain in potential energy of the block when it rises through the vertical
height h, and the net loss in potential energy of the system.
c. Using the value for friction found in Exercise 1, calculate the work done by
friction when the block moves the distance s up the plane.
d. Theoretically, the net loss in potential energy should equal the work done by
friction. Compare your results in parts (b) and (c) by calculating the percent deviation
between them.
4.
Prove that Eq. (5), which expresses the conservation of energy, is in fact the same
as Eq. (3a), which expresses the equilibrium of forces on a non-accelerating body.
6-3
```
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