20. Use the results from Example Problem 2 to answer questions abouta scale in an elevator on Earth. What force would be exerted bythe 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 2Real and Apparent Weight Your mass is 75.0 kg, and you arestanding on a bathroom scale in an elevator. Starting from rest,the elevator accelerates upward at 2.00 m/s² for 2.00 s and thencontinues at a constant speed. Is the scale reading during accelerationgreater than, equal to, or less than the scale reading when theelevator is at rest?1 Analyze and Sketch the Problem• Sketch the situation.F.scaleSystemChoose a coordinate system with the positivedirection as upward.F.netscale+yDraw the motion diagram. Label v and a.• Draw the free-body diagram. The net forceis in the same direction as the acceleration,so the upward force is greater than thedownward force.Known:Unknown:m= 75.0 kgE.scale?a = 2.00 m/s²t = 2.00 s= 9.80 N2 Solve for the UnknownFF.= manet+ (-F)F, is negative because it is in the negative direction defined bythe coordinate system.netscaleSolve for FscaleFnet + Fg= EscaleElevator at rest:F.= Eret + E.gThe elevator is not accelerating. Thus, FnetSubstitute Fet = 0.00 Nscalenet= 0.00 N.%3DgSubstitute F, = mgSubstitute m = 75.0 kg, g = 9.80 m/s²mg= (75.0 kg)(9.80 m/s²)= 735 NMath HandbookOperations withSignificant DigitsAcceleration of the elevator:pages 835–836Fecale = Fnet + g= ma + mg%3DSubstitute Fnet= ma, F, = mgSubstitute 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 NThe scale reading when the elevator is accelerating (885 N) is larger than the scalereading at rest (735 N).

Name: 20. Use the results from Example Problem 2 to answer questions abouta
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Description: 20. Use the results from Example Problem 2 to answer questions abouta scale in an elevator on Earth. What force would be exerted bythe scale on a person in the following situations?a. The elevator moves at constant speed.b. It slows at 2.00 m/s² whi

Given:- the mass of the person m = 75 kg Find:- The force exerted on the scale a) When the…

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- 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

- 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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ISBN:9781305952300

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

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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