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 AdRequired:Ad Ad AdYAnclysis:Let mooth & west be positive tv4d² = Ad + Ad2Sektion:N(20.0km) (sin 25°)(20.0xmXcos 25°)(45.0 kyksin 40) XWE(45.0km) (cos 40.00)40.0⁰Ad Ad + AdBd² = (-20.0kn) (co₁ 250) + (45.0 kn)(sin 40.0°)bota= 10.799 kmAd = A + ActAdy = (20.0 km.) (sin 250) + (45.0 kn X(cos 40.0")Ady = [42.924 km]| AZ²₂1 = √(40₂)² + (Ad4₂) ²142, 1 =√ √/(10.799 kn)² + (42.924 kn)²1Ad1=44.262 kmScheton (contid):-O + tan" (Ads)ở bán 42.921 km-110.799 Kn0 = 75.9°Statene tiThen fore, the displacement ofthe whole is 4.43 x 10¹ kn [W76″N MotionAd = Ad₁ + Ad₂ + ...bsin Basin AV₂ VtVAB = VAC + VCBa av==ForcesācF = ma=Csin Cc²=a² + b² sindtāav ( t)²2F = mã,Uniform Circular Motion1²F = may=-d = v₁ t+VXY =√xxā =v=x =FmF₂Unit 1 Dynamics=àtHcos-b+√b²-4ac2aF = mgGMmā=à = (v₁ + √₂) t2HV7.ttan 0 =1.AA==bhv² = ₁² +2αavF₁ = FNv=2-rV=Tc²=a² + b² - 2ab cos 0Y₂ Y₁X₂ X₁āav( t)²2āFA = lws,maxNSlope=d=v₂ ta = 4 ² rf² Tr=}V=FNGMr

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Description: 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 co

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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

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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