g=10 m/s^2 Frownμ=0.2030°II =Ffrict=Fav=420 Nkgm/s/sNa=ΣF=Fnorm (normal force):=Ffrict (force of friction)=m (mass) =kga (acceleration) =EF (net force) =NNm/s/sN

Answered: Image /qna-images/answer/39619cfc-fd26-42aa-9313-7c850fadf309.jpg

Answered: From μ=0.20 30° m= a= Ftrict= Fra=420 N…

From μ=0.20 30° m= a= Ftrict= Fra=420 N kg m/s/s N ΣF= Fnorm (normal force) = Ffrict (force of friction) = m (mass) = kg a (acceleration) = EF (net force) = N N m/s/s N

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter2: Representing Motion

Section: Chapter Questions

Problem 78A

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Question

g=10 m/s^2

Frown
μ=0.20
30°
II =
Ffrict=
Fav=420 N
kg
m/s/s
N
a=
ΣF=
Fnorm (normal force):
=
Ffrict (force of friction)
=
m (mass) =
kg
a (acceleration) =
EF (net force) =
N
N
m/s/s
N

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