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Recitation ClassWork 6 solutions
Physics 203
1.
Chapter 6 problem 2. In addition, draw an energy bar chart for each part. In part (a) choose the system of interest to
be the Earth, you, and your belongings. In part (b) choose the system of interest to be the Earth and you.
REASONING
a. We will use
( )
cos
WF
s
θ
=
(Equation 6.1) to determine the work
W
.
The force (magnitude =
F
) doing
the work is the force exerted on you by the elevator floor.
This force is the normal force and has a
magnitude
N
, so
=
N
in Equation 6.1.
To determine the normal force, we will use the fact that the
elevator is moving at a constant velocity and apply Newton’s second law with the acceleration set to zero.
Since the force exerted by the elevator and the displacement (magnitude =
s
) are in the same direction on
the upward part of the trip, the angle between them is
θ
= 0
°
, with the result that the work done by the force
is positive.
b.
To determine the normal force, we will again use the fact that the elevator is moving at a
constant velocity and apply Newton’s second law with the acceleration set to zero.
Since the force exerted by
the elevator and the displacement are in opposite directions on the downward part of the trip, the angle between
them is
θ
= 180
°
, and so the work done by the force is negative.
SOLUTION
a.
The freebody diagram at the right shows the three forces that
act on you:
W
is your weight,
W
b
is the weight of your belongings,
and
F
N
is the normal force exerted on you by the floor of the elevator.
Since you are moving upward at a constant velocity, your acceleration
is zero, you are in equilibrium, and the net force in the
y
direction
must be zero:
Nb
0
y
F
F WW
Σ
−−
=
(4.9b)
Therefore, the magnitude of the normal force is
F
N
=
W
+
W
b
.
The work done by the normal force is
( ) ( ) ( )
( )( )( )
cos
cos
685 N
915 N cos 0° 15.2 m
24 300 J
W F
s WW
s
θθ
=
=
+
=
+=
(6.1)
b.
During the downward trip, you are still in equilibrium since the elevator is moving with a
constant velocity.
The magnitude of the normal force is now
F
N
=
W
.
The work done by the normal force is
( ) ( ) ( )( )( )
N
cos
cos
685 N cos180° 15.2 m
10 400 J
W
F
sW
s
=
=
=
=
−
(6.1)
g
U
W
g
U
g
U
W
g
U
+y
F
N
W
W
b
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View Full Document 2.
Problem 9.
Hint: Careful when you determine the normal force exerted by the floor on the refrigerator.
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This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 RICHARDSON, B
 Physics, Energy, Work

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