Answer:
this is impossible for me
Explanation:
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.
In turn, we will resort to the application of Newton's second law.
PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

Where,
X = Desplazamiento
V = Velocity
t = Time
In this case there is no initial displacement or initial velocity, therefore

Clearing for time,



PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:
F = ma
Where,
m=mass
a = acceleration
Acceleration can also be written as,

Then





Negative symbol is because the force is opposite of the direction of moton.
PART C) Acceleration through kinematics equation is defined as




The gravity is equal to 0.8, then the acceleration is

