A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
s a distance of 2.7 m in 5.8 s. Find the coefficient of kinetic friction between the block and plane.
1 answer:
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
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Answer:
<h2>f₀ = 158.12 Hertz</h2>Explanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
where T is the tension in the string and
is the density of the string
Given T = 600N and = 0.015 g/cm = 0.0015kg/m
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:
where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:
Let us make mass, m, the subject of the formula:
From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore: