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HomeWork 7 solutions
Physics 203
1.
Chapter 6 problem 3. In part (b) the question is asking about the work done by the weight of the elevator. A better
way to say this is “the work done by the gravitational force exerted by the earth”. Also, the problem is not saying it
specifically, but the system of interest is the elevator (the Earth is not in the system).
REASONING
The work done by the tension in the cable is given by Equation 6.1 as
( )
cos
WT
s
θ
=
. Since
the elevator is moving upward at a constant velocity, it is in equilibrium, and the magnitude
T
of the tension
must be equal to the magnitude
mg
of the elevator’s weight;
T
=
mg
.
SOLUTION
a.
The tension and the displacement vectors point in the same direction (upward), so the angle
between them is
= 0
°
. The work done by the tension is
( ) ( )
( )( ) ( )
25
cos
cos
1200 kg 9.80 m/s
cos 0° 35 m
4.1 10 J
W
T
s
mg
s
θθ
=
=
=
=
×
(6.1)
b.
The weight and the displacement vectors point in opposite directions, so the angle between them
is
= 180
°
. The work done by the weight is
( ) ( )( ) ( )
cos
1200 kg 9.80 m/s
cos180° 35 m
4.1 10 J
W
mg
s
=
=
=
−×
(6.1)
2.
Problem 24. Before you do this problem, draw an energy bar chart for the situation.
I’ll choose the system to be just the skier. The initial moment will be when she is at the beginning
of the 21 m, and the final moment will be when she is at the end of the 21 m. Here’s the energy
bar chart. The skier starts with kinetic energy, but the friction force exerted by the snow on her
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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, Force, Work

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