0.90, draw a free body diagram representin

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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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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Kindly provide the solution to the following question using the GRASS method.  Free Body Diagrams and Problem Solving (Dynamics Unit): 

A Tim Horton’s Coffee Mug is moving along a level table towards the right as a result of having been pushed. If the Coffee Mug has a mass of 0.5 kg and the coefficient of friction between the coffee mug and the table is 0.90, draw a free body diagram representing all of the forces acting on the Coffee Mug.

 

 

Image attached is a guide on how to use the GRASS method on a question and the other image are the formulas for this unit.

Motion
Ad = Ad₁ + Ad₂ + ...
b
sin B
a
sin A
V₂ V
t
VAB = VAC + VCB
a av
=
=
Forces
āc
F = ma
=
C
sin C
c²=a² + b² sin
d
t
āav ( t)²
2
F = mã,
Uniform Circular Motion
1²
F = ma
y=-
d = v₁ t+
VXY =
√xx
ā =
v=
x =
F
m
F₂
Unit 1 Dynamics
=
à
t
H
cos
-b+√b²-4ac
2a
F = mg
GMm
ā=
à = (v₁ + √₂) t
2
H
V
7.
t
tan 0 =
1.
A
A==bh
v² = ₁² +2αav
F₁ = FN
v=2-r
V=
T
c²=a² + b² - 2ab cos 0
Y₂ Y₁
X₂ X₁
āav( t)²
2
ā
F
A = lw
s,max
N
Slope=
d=v₂ t
a = 4 ² rf² T
r=}
V=
FN
GM
r
Transcribed Image Text:Motion Ad = Ad₁ + Ad₂ + ... b sin B a sin A V₂ V t VAB = VAC + VCB a av = = Forces āc F = ma = C sin C c²=a² + b² sin d t āav ( t)² 2 F = mã, Uniform Circular Motion 1² F = ma y=- d = v₁ t+ VXY = √xx ā = v= x = F m F₂ Unit 1 Dynamics = à t H cos -b+√b²-4ac 2a F = mg GMm ā= à = (v₁ + √₂) t 2 H V 7. t tan 0 = 1. A A==bh v² = ₁² +2αav F₁ = FN v=2-r V= T c²=a² + b² - 2ab cos 0 Y₂ Y₁ X₂ X₁ āav( t)² 2 ā F A = lw s,max N Slope= d=v₂ t a = 4 ² rf² T r=} V= FN GM r
tiver:
Ad Ad
Required:
Ad Ad Ad
Y
Anclysis:
Let mooth & west be positive tv
4d² = Ad + Ad
2
Sektion:
N
(20.0km) (sin 25°)
(20.0xmXcos 25°)
(45.0 kyksin 40) X
WE
(45.0km) (cos 40.00)
40.0⁰
Ad Ad + Ad
Bd² = (-20.0kn) (co₁ 250) + (45.0 kn)(sin 40.0°)
bota
= 10.799 km
Ad = A + Act
Ady = (20.0 km.) (sin 250) + (45.0 kn X(cos 40.0")
Ady = [42.924 km]
| AZ²₂1 = √(40₂)² + (Ad4₂) ²
142, 1 =√ √/(10.799 kn)² + (42.924 kn)²
1Ad1=44.262 km
Scheton (contid):
-
O + tan" (Ads)
ở bán 42.921 km
-1
10.799 Kn
0 = 75.9°
Statene ti
Then fore, the displacement of
the whole is 4.43 x 10¹ kn [W76″N
Transcribed Image Text:tiver: Ad Ad Required: Ad Ad Ad Y Anclysis: Let mooth & west be positive tv 4d² = Ad + Ad 2 Sektion: N (20.0km) (sin 25°) (20.0xmXcos 25°) (45.0 kyksin 40) X WE (45.0km) (cos 40.00) 40.0⁰ Ad Ad + Ad Bd² = (-20.0kn) (co₁ 250) + (45.0 kn)(sin 40.0°) bota = 10.799 km Ad = A + Act Ady = (20.0 km.) (sin 250) + (45.0 kn X(cos 40.0") Ady = [42.924 km] | AZ²₂1 = √(40₂)² + (Ad4₂) ² 142, 1 =√ √/(10.799 kn)² + (42.924 kn)² 1Ad1=44.262 km Scheton (contid): - O + tan" (Ads) ở bán 42.921 km -1 10.799 Kn 0 = 75.9° Statene ti Then fore, the displacement of the whole is 4.43 x 10¹ kn [W76″N
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