Respuesta :
Answer:
a. 25 N/m
b. 0.8886 s
c. 0.707 m/s
d. At the equilibrium point
Explanation:
m = 500 g = 0.5 kg
L = 20 cm = 0.2 m
A = 10 cm = 0.1 m
a. Let g = 10 m/s2, then the gravity of the 0.5 kg book acting on the spring is
F = mg = 0.5*10 = 5 N
If the spring is stretched L = 0.2 m under 5N load, then the spring constant k is:
k = F/l = 5 / 0.2 = 25 N/m
b. We can treat this as simple harmonic motion with magnitude A = 0.1 cm. The period of this motion is
[tex]T = 2\pi \sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{25}} = 0.8886 s[/tex]
c. The book maximum speed:
[tex]\omega A = \sqrt{\frac{k}{m}}A = \sqrt{\frac{25}{0.5}}0.1 = 0.707 m/s[/tex]
d. Due to the law of energy conservation, the maximum speed would occur at the equilibrium point. This is where the potential energy, elastic energy is 0 and the kinetic energy is greatest.
Newton's second law and the relationships of simple harmonic motion allows to find the results for questions about the motion of mass-spring are:
a) The spring constant is: k = 24.5 N / m
b) The period is: T = 0.128 s
c) the maximum velocity is v = 4.9 m / s and occurs in the equilibrium position.
Given parameters
- The mass of the body m = 500g = 0.500 kg
- The elongation of the spring x = 20 cm = 0.20 cm
To find
a) The spring constant.
b) It lengthens x = 10 cm = 0.10 m from equilibrium. What is the angular velocity.
c) The maximum speed and in what position it occurs.
Newton's second law gives a relationship between the net force, the mass, and the acceleration of the body.
Σ F = m a
Where m is the mass and the acceleration of the bodies.
Hooke's law states that the force of a spring is proportional to the displacement.
F = - k x
Where k is the spring constant and x is the displacement from the equilibrium position.
a) In the attached we have a free-body diagram of the system, we used the Newton's second law.
F-W = 0
F = W
k x = mg
k = [tex]\frac{mg}{x}[/tex]mg / x
Let's calculate
k = [tex]\frac{0.500 \ 9.8}{0.20}[/tex]
k = 24.5 N / m
b) In simple harmonic motion the expression that describes it is:
x = A cos (wt + Ф)
w² = k / m
Where A is the extra elongation or amplitude, w the angular velocity, t the time and Ф a phase constant that is determined with the initial conditions.
w = [tex]\sqrt{\frac{24.5}{0.500} }[/tex]Ra 24.5 / 0.500
w = 49 rad / s
The angular velocity and period are related
[tex]w= \frac{2\pi }{T} \\T= \frac{2\pi }{w} \\ \\ T = \frac{2\pi }{49}[/tex]
T = 0.128 s
c) The speed of defined by the change of position in the unit of time
v = [tex]\frac{dx}{dt}[/tex] t
v = -A w sin (wt + Ф)
Elongation of motion is the stretch from the starting position that was in balance with the weight.
a = 0.10 m
The speed is maximum when the sinus function is maximum.
sin θ = ± 1
v = Aw
v = 0.10 49
v = 4.9 m / s
To find in which position it occurs we use that when the sine is maximum, the cosine is zero, therefore the position is;
x = 0
This is the equilibrium position around which the 10 cm elongation occurs.
In conclusion using Newton's second law and the simple harmonic motion relationships we can find the results for the questions about the motion of mass-spring are:
a) The spring constant is: k = 24.5 N / m
b) The period is: T = 0.128 s
c) the maximum velocity is v = 4.9 m / s and occurs in the equilibrium position.
Learn more here: brainly.com/question/21807111
