Questión 2 In the situation illustrated in the figure below, a small package is deposited by a conveyor belt at point A onto a ramp at e = 20° degrees to the horizontal with a velocity of (0.5 m/s. The distance between points A and Bis 2m. B a) Calculate the maximum value u of the kinetic friction coefficient between the package and the ramp AB for which the parcel reaches point B. b) If the actual kinetic friction coefficient between the package and AB and BC is µ4 = 0.5/43, calculate the distance s on the level surface BC at which the package comes to rest.
Questión 2 In the situation illustrated in the figure below, a small package is deposited by a conveyor belt at point A onto a ramp at e = 20° degrees to the horizontal with a velocity of (0.5 m/s. The distance between points A and Bis 2m. B a) Calculate the maximum value u of the kinetic friction coefficient between the package and the ramp AB for which the parcel reaches point B. b) If the actual kinetic friction coefficient between the package and AB and BC is µ4 = 0.5/43, calculate the distance s on the level surface BC at which the package comes to rest.
Physics for Scientists and Engineers: Foundations and Connections
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ISBN:9781133939146
Author:Katz, Debora M.
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Chapter5: Newton's Laws Of Motion
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Problem 74PQ: Starting from rest, a rectangular toy block with mass 300 g slides in 1.30 s all the way across a...
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![Question 2
In the situation illustrated in the figure below, a small package is deposited by a conveyor belt at point A onto a
ramp at e = 20° degrees to the horizontal with a velocity of (0.5 m/s. The distance between points A and Bis
2 m.
C
B
a)
Calculate the maximum value p of the kinetic friction coefficient between the package and the ramp AB
for which the parcel reaches point B.
b)
If the actual kinetic friction coefficient between the package and AB and BC is uk = 0.54, calculate the
distance s on the level surface BC at which the package comes to rest.
Note: if you haven't been able to solve part a), take the actual kinetic friction coefficient equal to us = 0.15.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d8b7bee-36d5-492d-920f-c603eda04374%2F093948e8-a649-4a20-8ca7-5869b0c7d4ac%2Fur8mb3q_processed.png&w=3840&q=75)
Transcribed Image Text:Question 2
In the situation illustrated in the figure below, a small package is deposited by a conveyor belt at point A onto a
ramp at e = 20° degrees to the horizontal with a velocity of (0.5 m/s. The distance between points A and Bis
2 m.
C
B
a)
Calculate the maximum value p of the kinetic friction coefficient between the package and the ramp AB
for which the parcel reaches point B.
b)
If the actual kinetic friction coefficient between the package and AB and BC is uk = 0.54, calculate the
distance s on the level surface BC at which the package comes to rest.
Note: if you haven't been able to solve part a), take the actual kinetic friction coefficient equal to us = 0.15.
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Follow-up Question
This is how I did it. Where did I go wrong?
![(A)
Foom
A
to B
VA =O m/Sec
VB= 0.5m/sec
SA-B= am
0= 20
Velocity and accelesation
eqLation
2
0.5) co)² &xaix2
a1=0.0625 0/5?
AT Block
W mg
\20
a
Fs =lkN
J20
NEmg Cosao
Efx= man
maSingi- MKN =mai
my Sinai -uk mg Cos2o) =m ai
.: 9,=g (Sindo- Uk Co saó
O'0625= 9.81( Sin2o -Uk Cos)](https://content.bartleby.com/qna-images/question/2d8b7bee-36d5-492d-920f-c603eda04374/7de2b921-3bff-47d1-9a1f-413344db1b92/hinq4f_thumbnail.jpeg)
Transcribed Image Text:(A)
Foom
A
to B
VA =O m/Sec
VB= 0.5m/sec
SA-B= am
0= 20
Velocity and accelesation
eqLation
2
0.5) co)² &xaix2
a1=0.0625 0/5?
AT Block
W mg
\20
a
Fs =lkN
J20
NEmg Cosao
Efx= man
maSingi- MKN =mai
my Sinai -uk mg Cos2o) =m ai
.: 9,=g (Sindo- Uk Co saó
O'0625= 9.81( Sin2o -Uk Cos)
![uK= 0.357
uk)ad=
0.5x 0.357e 0.1785
Between
B to
o c
at
rest
Vc-om/sec
:-
=VB?+ 292S
2
mg
AT Block
N=mg
Ef= man
-maz + lk.N=0
12
-Mk (Mg.)
- mar
108 m/2
42=-ukg - -0.1185x9.81= -1.15
0= (0.5)+ 2x - 1.75168 x (S)
S = 0.07138m](https://content.bartleby.com/qna-images/question/2d8b7bee-36d5-492d-920f-c603eda04374/7de2b921-3bff-47d1-9a1f-413344db1b92/y338v8a_thumbnail.jpeg)
Transcribed Image Text:uK= 0.357
uk)ad=
0.5x 0.357e 0.1785
Between
B to
o c
at
rest
Vc-om/sec
:-
=VB?+ 292S
2
mg
AT Block
N=mg
Ef= man
-maz + lk.N=0
12
-Mk (Mg.)
- mar
108 m/2
42=-ukg - -0.1185x9.81= -1.15
0= (0.5)+ 2x - 1.75168 x (S)
S = 0.07138m
Solution
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