Determine the acceleration of a 55 kg sled coming down a snow hill at 30 degrees to the horizontal ground if the coefficient of friction between the sled and the snow hill is 0.14.

Physics for Scientists and Engineers, Technology Update (No access codes included)
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ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter15: Oscillatory Motion
Section: Chapter Questions
Problem 15.12OQ: For a simple harmonic oscillator, answer yes or no to the following questions, (a) Can the...
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Kindly provide the solution to the following question using the GRASS method.  Free Body Diagrams and Problem Solving (Dynamics Unit): 

Determine the acceleration of a 55 kg sled coming down a snow hill at 30 degrees to the horizontal ground if the coefficient of friction between the sled and the snow hill is 0.14.

 

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

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
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
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