Suppose that a spring is attached to a wall. On the other end of the spring, it is attached to an object of 0.263kg. When the spring is compressed at a distance of 15cm, it holds an elastic potential energy of 20.0J. The horizontal surface has no friction; frictionless.

a) If the motion has an amplitude of 7.00cm and a period of 3.0s, where is the object and in what direction is it moving 0.45 seconds after it has passed the equilibrium position, moving to the left?
b) calculate the system's KE and PE In the moment expressed in part b
c) What force ( magnitude and direction) does the spring apply on the object when it is 3.50cm to the right of the equilibrium position, moving to the right?

To solve this problem, we need to use the concepts of spring potential energy, simple harmonic motion, and forces. Let's break down the problem step by step:

a) To determine the position and direction of the object 0.45 seconds after it has passed the equilibrium position, moving to the left, we can use the equation for simple harmonic motion:
x(t) = A * cos(2πt / T)
Where:
x(t) is the object's position at time t
A is the amplitude of the motion
T is the period of the motion

Given:
A = 7.00 cm = 0.07 m (converting from cm to m)
T = 3.0 s

To find the position at 0.45 seconds after passing the equilibrium position:
t = 0.45 s

Substituting the values into the equation:
x(0.45) = 0.07 * cos(2π * 0.45 / 3)

Calculating:

x(0.45) = 0.07 * cos(0.94248)

Using a calculator, we find:

x(0.45) ≈ 0.070 m

Therefore, the object is located 0.070 meters to the left of the equilibrium position.

b) To calculate the system's kinetic energy (KE) and potential energy (PE) at this moment, we need to use the following equations:

KE = 0.5 * m * v^2
PE = 0.5 * k * x^2

Given:
m = 0.263 kg (mass of the object)
x = 0.070 m (position of the object)

The velocity (v) can be found using the equation for simple harmonic motion:
v(t) = -A * 2π / T * sin(2πt / T)

At 0.45 seconds after passing the equilibrium position:
t = 0.45 s

Substituting the values into the equation:
v(0.45) = -0.07 * 2π / 3 * sin(2π * 0.45 / 3)

Calculating:

v(0.45) = -0.07 * 2π / 3 * sin(0.94248)

Using a calculator, we find:

v(0.45) ≈ -0.184 m/s

Now, calculating KE and PE:

KE = 0.5 * 0.263 * (-0.184)^2
PE = 0.5 * k * (0.070)^2

Given:
Elastic potential energy (PE) = 20.0 J

We can equate the elastic potential energy (PE) to the potential energy at this position (PE) calculated above:

PE = 20.0 J

Substituting the values:

20.0 J = 0.5 * k * (0.070)^2

Solving for k:

k = (20.0 J) / (0.5 * (0.070)^2)

Calculating:

k ≈ 877.55 N/m

Now, calculating KE and PE:

KE = 0.5 * 0.263 * (-0.184)^2
PE = 0.5 * 877.55 * (0.070)^2

Calculating:

KE ≈ 0.002264 J
PE ≈ 0.1724 J

Therefore, at this moment, the system's kinetic energy (KE) is approximately 0.002264 J, and the potential energy (PE) is approximately 0.1724 J.

c) To determine the force exerted by the spring when the object is 3.50 cm to the right of the equilibrium position, moving to the right, we can use Hooke's Law:

F = -k * x

Given:
k = 877.55 N/m (spring constant)
x = 3.50 cm = 0.035 m (converting from cm to m)

Substituting the values into the equation:

F = -877.55 N/m * 0.035 m

Calculating:

F ≈ -30.716 N

Therefore, the force exerted by the spring is approximately 30.716 N to the left.

a) To determine the position and direction of the object at a specific time, we can use the equation for simple harmonic motion: x(t) = A * cos(ωt + φ), where x(t) is the position of the object at time t, A is the amplitude (maximum displacement), ω is the angular frequency (2π / T, where T is the period), and φ is the phase constant.

Given:
Amplitude (A) = 7.00 cm = 0.07 m
Period (T) = 3.0 s
Time (t) = 0.45 s
Direction = moving to the left

First, find the angular frequency:
ω = 2π / T = 2π / 3.0 = 2.094 s⁻¹

Substitute the values into the equation:
x(t) = 0.07 * cos(2.094 * 0.45 + φ)

To determine the phase constant (φ), we need an initial condition. Let's assume that at t = 0 s, the object is at the equilibrium position (x = 0). This implies that cos(φ) = 0, which means φ = π/2.

Replace φ with π/2:
x(t) = 0.07 * cos(2.094 * 0.45 + π/2)

Calculate the expression:
x(t) = 0.07 * cos(1.939 + π/2)
x(t) = 0.07 * cos(2.477)

Evaluate the cosine using a calculator:
x(t) = 0.07 * (-0.781) ≈ -0.0547 m

Therefore, 0.45 seconds after passing the equilibrium position and moving to the left, the object is located approximately -0.0547 m (or 5.47 cm) to the left of the equilibrium position.

b) To calculate the kinetic energy (KE) and potential energy (PE) of the system at this moment, we'll use the formulas:

KE = 0.5 * m * v^2 (where m is the mass and v is the velocity)
PE = 0.5 * k * x^2 (where k is the spring constant and x is the displacement from equilibrium)

Given:
Mass (m) = 0.263 kg
Velocity (v) = ? (We need to find it)
Spring constant (k) = ?
Displacement (x) = -0.0547 m

To find the velocity, we can differentiate the equation for simple harmonic motion:
v(t) = -A * ω * sin(ωt + φ)

Substituting values:
v(t) = -0.07 * 2.094 * sin(2.094 * 0.45 + π/2)

Evaluate the expression using a calculator:
v(t) = -0.07 * 2.094 * sin(1.939 + π/2)
v(t) ≈ -0.07 * 2.094 * (-0.623) ≈ 0.0925 m/s

Now we can calculate the kinetic energy:
KE = 0.5 * m * v^2
KE = 0.5 * 0.263 * (0.0925)^2

Calculate the expression:
KE ≈ 0.006054 J

To find the spring constant (k), we can use the equation for elastic potential energy:
PE = 0.5 * k * x^2

Rearrange the equation to solve for k:
k = 2 * PE / x^2
k = 2 * 20.0 / (-0.0547)^2

Calculate the expression:
k ≈ 15879.1 N/m

Finally, calculate the potential energy:
PE = 0.5 * k * x^2
PE = 0.5 * 15879.1 * (-0.0547)^2

Calculate the expression:
PE ≈ 22.7 J

Therefore, the system's kinetic energy (KE) at this moment is approximately 0.00605 J, and its potential energy (PE) is approximately 22.7 J.

c) To determine the force exerted by the spring on the object when it is 3.50 cm to the right of the equilibrium position and moving to the right, we can use Hooke's Law:

F = -k * x

Given:
Spring constant (k) = 15879.1 N/m
Displacement (x) = 0.035 m (3.50 cm)
Direction = moving to the right

Substituting the values into the equation:
F = -15879.1 * 0.035
F ≈ -555.2 N

The negative sign indicates that the force is in the opposite direction to the displacement. Therefore, the spring applies a force of approximately 555.2 N to the left on the object when it is 3.50 cm to the right of the equilibrium position and moving to the right.