You attach one end of a spring with a force constant

k = 893 N/m
to a wall and the other end to a mass
m = 2.02 kg
and set the mass-spring system into oscillation on a horizontal frictionless surface as shown in the figure. To put the system into oscillation, you pull the block to a position
xi = 6.56 cm
from equilibrium and release it.

a) Determine the potential energy stored in the spring before the block is released.

(b) Determine the speed of the block as it passes through the equilibrium position.

(c) Determine the speed of the block when it is at a position
xi/4.

a) The potential energy stored in the spring before the block is released is U = 1/2 kx^2 = 1/2 (893 N/m)(0.0656 m)^2 = 3.7 J.

b) The speed of the block as it passes through the equilibrium position is v = sqrt(2U/m) = sqrt(2(3.7 J)/(2.02 kg)) = 1.6 m/s.

c) The speed of the block when it is at a position xi/4 is v = sqrt(2U/m) = sqrt(2(3.7 J)/(2.02 kg)) = 1.6 m/s.

a) You want to determine the potential energy stored in the spring before the block is released.

The potential energy stored in a spring is given by the formula:

Potential energy = (1/2) * k * x^2

where k is the force constant and x is the displacement from the equilibrium position.

Given:
k = 893 N/m
x = 6.56 cm = 0.0656 m (convert cm to m)

Plugging in the values into the formula, we get:

Potential energy = (1/2) * (893 N/m) * (0.0656 m)^2

Simplifying the equation, we get:

Potential energy = 0.5 * 893 * 0.00431296

Potential energy = 1.88074 J (rounded to 5 decimal places)

So, the potential energy stored in the spring before the block is released is approximately 1.88074 J.

b) To find the speed of the block as it passes through the equilibrium position, we can use the principle of conservation of mechanical energy.

The total mechanical energy of the system is the sum of the potential energy and kinetic energy.

Potential energy at equilibrium position = (1/2) * k * x^2 = (1/2) * 893 * (0 m)^2 = 0 J

Total mechanical energy = Potential energy + Kinetic energy

When the block passes through the equilibrium position, all the potential energy is converted into kinetic energy. Therefore, the total mechanical energy is equal to the kinetic energy.

Since mechanical energy is conserved, we have:

Total mechanical energy = Initial potential energy = 1.88074 J

Kinetic energy = Total mechanical energy - Potential energy at equilibrium position = 1.88074 J - 0 J = 1.88074 J

To find the speed of the block, we can use the formula for kinetic energy:

Kinetic energy = (1/2) * m * v^2

Rearranging the equation, we get:

v^2 = (2 * Kinetic energy) / m

Plugging in the values, we get:

v^2 = (2 * 1.88074 J) / 2.02 kg

v^2 = 1.86173168168 m^2/s^2

Taking the square root of both sides, we get:

v = sqrt(1.86173168168) m/s

v ≈ 1.36 m/s (rounded to 2 decimal places)

So, the speed of the block as it passes through the equilibrium position is approximately 1.36 m/s.

c) To find the speed of the block when it is at a position xi/4, we can use the principle of conservation of mechanical energy again.

The potential energy at a position xi/4 can be found using the following formula:

Potential energy at position xi/4 = (1/2) * k * (xi/4)^2

Given:
k = 893 N/m
xi = 6.56 cm = 0.0656 m (convert cm to m)

Plugging in the values into the formula, we get:

Potential energy at position xi/4 = (1/2) * 893 * (0.0656/4)^2

Simplifying the equation, we get:

Potential energy at position xi/4 = 0.5 * 893 * 0.000425984

Potential energy at position xi/4 = 0.19053 J (rounded to 5 decimal places)

Now, let's find the total mechanical energy at position xi/4.

Total mechanical energy = Potential energy at position xi/4 + Kinetic energy

Since mechanical energy is conserved, the total mechanical energy at position xi/4 is equal to the initial potential energy:

Total mechanical energy at position xi/4 = Initial potential energy = 1.88074 J

To find the kinetic energy at position xi/4, we can subtract the potential energy at that position from the total mechanical energy:

Kinetic energy at position xi/4 = Total mechanical energy at position xi/4 - Potential energy at position xi/4 = 1.88074 J - 0.19053 J = 1.69021 J

Now, we can find the speed of the block using the formula for kinetic energy:

Kinetic energy = (1/2) * m * v^2

Rearranging the equation, we get:

v^2 = (2 * Kinetic energy) / m

Plugging in the values, we get:

v^2 = (2 * 1.69021 J) / 2.02 kg

v^2 = 1.6742663398 m^2/s^2

Taking the square root of both sides, we get:

v = sqrt(1.6742663398) m/s

v ≈ 1.29 m/s (rounded to 2 decimal places)

So, the speed of the block when it is at a position xi/4 is approximately 1.29 m/s.

(a) To determine the potential energy stored in the spring before the block is released, we can use the formula for potential energy in a spring:

Potential energy = (1/2) * k * (x^2)

Where k is the force constant of the spring and x is the displacement from equilibrium.

Given:
k = 893 N/m
x = 6.56 cm = 0.0656 m

Plugging in these values into the formula:

Potential energy = (1/2) * 893 N/m * (0.0656 m)^2
= 2.11 J

So, the potential energy stored in the spring before the block is released is 2.11 J.

(b) To determine the speed of the block as it passes through the equilibrium position, we can use the principle of conservation of mechanical energy. At the equilibrium position, all the potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at release to the kinetic energy at the equilibrium position.

Potential energy at release = Kinetic energy at equilibrium position

Potential energy at release = 2.11 J

Kinetic energy at equilibrium position = (1/2) * m * v^2

Given:
m = 2.02 kg

Plugging in the values and solving for v:

2.11 J = (1/2) * 2.02 kg * v^2
v^2 = (2 * 2.11 J) / 2.02 kg
v^2 = 2.21 m^2/s^2

Taking the square root of both sides:

v = √(2.21) m/s
v ≈ 1.49 m/s

So, the speed of the block as it passes through the equilibrium position is approximately 1.49 m/s.

(c) To determine the speed of the block when it is at a position xi/4, we can use the principle of conservation of mechanical energy. We can equate the initial potential energy to the sum of the potential energy and kinetic energy at the position xi/4.

Potential energy at release = Potential energy at xi/4 + Kinetic energy at xi/4

Potential energy at release = 2.11 J

Potential energy at xi/4 = (1/2) * k * (xi/4)^2

Given:
xi/4 = 6.56 cm / 4 = 0.0164 m

Plugging in the values and solving for the potential energy at xi/4:

Potential energy at xi/4 = (1/2) * 893 N/m * (0.0164 m)^2
= 0.022 J

So, the potential energy at xi/4 is approximately 0.022 J.

Using the principle of conservation of mechanical energy:

2.11 J = 0.022 J + (1/2) * m * v'^2

Rearranging the equation and solving for v':

v'^2 = (2.11 J - 0.022 J) / (1/2) * 2.02 kg
v'^2 ≈ 2.09 m^2/s^2

Taking the square root of both sides:

v' ≈ √(2.09) m/s
v' ≈ 1.45 m/s

So, the speed of the block when it is at a position xi/4 is approximately 1.45 m/s.

To solve these questions, we need to apply the principles of potential energy and conservation of mechanical energy. We'll break down each part of the problem step by step.

(a) Determine the potential energy stored in the spring before the block is released:

The potential energy stored in a spring can be calculated using the formula:

Potential Energy (PE) = (1/2)kx^2

Where k is the force constant of the spring and x is the displacement from equilibrium.

Given:
k = 893 N/m
x = 6.56 cm = 0.0656 m

Substituting these values into the formula:

PE = (1/2) * 893 * (0.0656)^2
PE ≈ 1.4503 J

So, the potential energy stored in the spring before the block is released is approximately 1.4503 Joules.

(b) Determine the speed of the block as it passes through the equilibrium position:

To determine the speed of the block when it passes through the equilibrium position, we can use the principle of conservation of mechanical energy. At the equilibrium position, all potential energy is converted into kinetic energy.

The potential energy at the equilibrium position is zero (as the spring is not stretched or compressed). Therefore, all the initial potential energy is converted into kinetic energy.

Kinetic Energy (KE) = Potential Energy (PE)

Let's denote the speed of the block at the equilibrium position as v_eq.

We can write the equation:

(1/2) * m * (v_eq)^2 = PE

Substituting the given values:

(1/2) * 2.02 * (v_eq)^2 = 1.4503

Solving for v_eq:

(v_eq)^2 ≈ (1.4503 * 2) / 2.02
(v_eq)^2 ≈ 1.4361
v_eq ≈ √1.4361
v_eq ≈ 1.198 m/s

So, the speed of the block as it passes through the equilibrium position is approximately 1.198 m/s.

(c) Determine the speed of the block when it is at a position xi/4:

To determine the speed of the block at a position xi/4, we can use the principle of conservation of mechanical energy again. At any position, the total mechanical energy (potential energy + kinetic energy) remains constant.

Let's denote the speed of the block at position xi/4 as v_xi/4.

The total mechanical energy at position xi/4 is equal to the initial potential energy:

Total Mechanical Energy = Initial Potential Energy

Let's denote the displacement at position xi/4 as x_xi/4.

We can write the equation:

(1/2) * k * (x_xi/4)^2 + (1/2) * m * (v_xi/4)^2 = PE

We know that x_xi/4 = (6.56 cm) / 4 = 1.64 cm = 0.0164 m

Substituting the given values:

(1/2) * 893 * (0.0164)^2 + (1/2) * 2.02 * (v_xi/4)^2 = 1.4503

Solving for v_xi/4:

(v_xi/4)^2 ≈ [(1.4503 - 0.0164^2 * 893) * 2] / 2.02
(v_xi/4)^2 ≈ 1.0158
v_xi/4 ≈ √1.0158
v_xi/4 ≈ 1.008 m/s

So, the speed of the block when it is at a position xi/4 is approximately 1.008 m/s.