Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
<u>Answer:</u>
2N/cm
<u>Step-by-step explanation:</u>
According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:
where, 
is the force which is stretching or compressing the spring,
is the spring constant; and
is the distance the spring is stretched.
Substituting the given values to find the elastic constant to get:
Therefore, the elastic constant is 2 Newton/cm.
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
