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Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = ( × mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s
Answer:
F = 233.52 N, θ' = 351.41º
Explanation:
In this exercise we must find the net force applied on the donkey.
For this we use Newton's second law, where we create a reference frame with the horizontal x axis
let's decompose the forces
Jack
= 80.5 N
Jill
cos 45 = F_{2x} / F₂2
sin 45 = F_{2y} / F₂2
F_{2x} = F₂ cos 45
F_{2y} = F₂ sin 45
F_{2x} = 81.7 cos 45 = 57.77 N
F_{2y} = 81.7 sin 45 = 57.77 N
Jane
cos (270 + 45) = F_{3x} / F₃3
sin 315 = F_{3y} / F₃
F_{3x} = 131 cos 315 = 92.63 N
F_{3y} = 131 sin 315 = -92.63 N
the force can be found in each axis
X axis
F_{x} = F_{1x} + F_{2x} + F_{3x}
F_{x} = 80.5 +57.77 + 92.63
F_{x} = 230.9 N
Axis y
F_{y} = F_{1y} + F_{2y} + F_{3y}
F_{y} = 0 + 57.77 -92.63
F_{y} = -34.86 N
we can give the result in two ways
a) F = (230.9 i ^ - 34.86 j ^) N
b) in the form of module and angle
we use the Pythagorean theorem
F = √(Fₓ² + F_{y}²
F = √(230.9² + 34.86²)
F = 233.52 N
let's use trigonometry for the angle
tan θ =
θ = tan⁻¹ (\frac{F_y}{F_x} })
θ = tan⁻¹ (-34.86 / 230.9)
θ = -8.59º
if we measure this angle from the positive side of the x-axis counterclockwise
θ' = 360 -θ
θ‘= 360- 8.59
θ' = 351.41º