Respuesta :
Answer:
μ = 0.423
Explanation:
To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body
Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis
x = v₀ t + ½ a t²
The body starts from rest so its initial speed is zero
a = 2 x / t²
a = 2 0.5 /0.5²
a = 4 m / s²
Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive
X axis
fr - Wₓ = m a (1)
Y Axis
N- [tex]W_{y}[/tex] = 0
N = W_{y}
We use trigonometry to find the components of the weight
sin 45 = Wₓ / W
cos 45 = W_{y} / W
Wₓ = W sin 45
W_{y} = W cos 45
The out of touch has the expression
fr = μ N
fr = μ W_{y}
We substitute in 1
μ mg cos 45 - mg sin 45 = m a
W_{y} = (a + g sin 45) / g cos 45
μ = a / g cos 45 + 1
We calculate
Acceleration goes down the plane, so it is negative
a = -4 m / s²
μ = 1- 4 / (9.8 cos 45)
μ = 0.423
Answer:
The μ = 0.422
Explanation:
The distance travelled by the mass is equal to:
[tex]L=ut+\frac{1}{2}at^{2} \\0.5=(0*5)+\frac{1}{2} a(0.5^{2} )\\a=4m/s^{2}[/tex]
The sum of forces in y-direction equals zero:
∑Fy = 0
N - (m * g * cosθ) = 0
N - (1 * 9.8 * cos45) = 0
N = 6.93 N
The sum of forces in x-direction is equal to:
∑Fx = ma
(m * g * sinθ) - fk = m * a
(1 * 9.8 * sin45) - fk = 1 * 4
fk = 2.93 N
fk = μ * N
2.93 = μ * 6.93
μ = 0.422
