- Use the results from Example Problem 2 to ansv a scale in an elevator on Earth. What force woul the scale on a person in the following situations a. The elevator moves at constant speed. b. It slows at 2.00 m/s² while moving upward. c. It speeds up at 2.00 m/s² while moving dowr movee downuword ot oonotont onoed

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20. Use the results from Example Problem 2 to answer questions about
a scale in an elevator on Earth. What force would be exerted by
the scale on a person in the following situations?
a. The elevator moves at constant speed.
b. It slows at 2.00 m/s² while moving upward.
c. It speeds up at 2.00 m/s² while moving downward.
d. It moves downward at constant speed.
e. It slows to a stop at a constant magnitude of acceleration.
Transcribed Image Text:20. Use the results from Example Problem 2 to answer questions about a scale in an elevator on Earth. What force would be exerted by the scale on a person in the following situations? a. The elevator moves at constant speed. b. It slows at 2.00 m/s² while moving upward. c. It speeds up at 2.00 m/s² while moving downward. d. It moves downward at constant speed. e. It slows to a stop at a constant magnitude of acceleration.
EXAMPLE Problem 2
Real and Apparent Weight Your mass is 75.0 kg, and you are
standing on a bathroom scale in an elevator. Starting from rest,
the elevator accelerates upward at 2.00 m/s² for 2.00 s and then
continues at a constant speed. Is the scale reading during acceleration
greater than, equal to, or less than the scale reading when the
elevator is at rest?
1 Analyze and Sketch the Problem
• Sketch the situation.
F.
scale
System
Choose a coordinate system with the positive
direction as upward.
F.
net
scale
+y
Draw the motion diagram. Label v and a.
• Draw the free-body diagram. The net force
is in the same direction as the acceleration,
so the upward force is greater than the
downward force.
Known:
Unknown:
m= 75.0 kg
E.
scale
?
a = 2.00 m/s²
t = 2.00 s
= 9.80 N
2 Solve for the Unknown
F
F.
= ma
net
+ (-F)
F, is negative because it is in the negative direction defined by
the coordinate system.
net
scale
Solve for F
scale
Fnet + Fg
= E
scale
Elevator at rest:
F.
= Eret + E.
g
The elevator is not accelerating. Thus, Fnet
Substitute Fet = 0.00 N
scale
net
= 0.00 N.
%3D
g
Substitute F, = mg
Substitute m = 75.0 kg, g = 9.80 m/s²
mg
= (75.0 kg)(9.80 m/s²)
= 735 N
Math Handbook
Operations with
Significant Digits
Acceleration of the elevator:
pages 835–836
Fecale = Fnet + g
= ma + mg
%3D
Substitute Fnet
= ma, F, = mg
Substitute m = 75.0 kg, a = 2.00 m/s², g = 9.80 m/s²
= m(a + g)
(75.0 kg)(2.00 m/s² + 9.80 m/s²)
= 885 N
The scale reading when the elevator is accelerating (885 N) is larger than the scale
reading at rest (735 N).
Transcribed Image Text:EXAMPLE Problem 2 Real and Apparent Weight Your mass is 75.0 kg, and you are standing on a bathroom scale in an elevator. Starting from rest, the elevator accelerates upward at 2.00 m/s² for 2.00 s and then continues at a constant speed. Is the scale reading during acceleration greater than, equal to, or less than the scale reading when the elevator is at rest? 1 Analyze and Sketch the Problem • Sketch the situation. F. scale System Choose a coordinate system with the positive direction as upward. F. net scale +y Draw the motion diagram. Label v and a. • Draw the free-body diagram. The net force is in the same direction as the acceleration, so the upward force is greater than the downward force. Known: Unknown: m= 75.0 kg E. scale ? a = 2.00 m/s² t = 2.00 s = 9.80 N 2 Solve for the Unknown F F. = ma net + (-F) F, is negative because it is in the negative direction defined by the coordinate system. net scale Solve for F scale Fnet + Fg = E scale Elevator at rest: F. = Eret + E. g The elevator is not accelerating. Thus, Fnet Substitute Fet = 0.00 N scale net = 0.00 N. %3D g Substitute F, = mg Substitute m = 75.0 kg, g = 9.80 m/s² mg = (75.0 kg)(9.80 m/s²) = 735 N Math Handbook Operations with Significant Digits Acceleration of the elevator: pages 835–836 Fecale = Fnet + g = ma + mg %3D Substitute Fnet = ma, F, = mg Substitute m = 75.0 kg, a = 2.00 m/s², g = 9.80 m/s² = m(a + g) (75.0 kg)(2.00 m/s² + 9.80 m/s²) = 885 N The scale reading when the elevator is accelerating (885 N) is larger than the scale reading at rest (735 N).
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