15 points pls help will give the crown to best answer 15 points!!!
2 answers:
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Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
Answer:
D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.
Explanation:
Newton's third law
Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.
F₁₂= -F₂₁
F₁₂: Force of the box on the worker
F₂₁: Force of the worker on the box
Newton's second law
∑F = m*a
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Formula to calculate the mass (m)
m = W/g
Where:
W : Weight (N)
g : acceleration due to gravity (m/s²)
Data
W₁ =1.8 kN : box weight
W₂ = 0.900 kN : worker weight
F₂₁ = 0.200 kN
F₁₂ = - 0.200 kN
g = 9.8 m/s²
Newton's second law for the box
∑F = m*a
F₂₁ = m₁*a₁ m₁=W₁/g
0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁
a₁= 1.08 m/s² : acceleration of the box
Newton's second law for the worker
∑F = m*a
F₁₂ = m₂*a₂ , m₂=W₂/g
- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂
a₂= -2.17 m/s² : acceleration of the worker