A 3.5 kg object is suspended from a spring of spring constant 725 N/m. Starting from the equilibrium position, the object has an initial speed of 7.9 m/s. What is the maximum displacement of the object during its subsequent motion? (In cm)

Kinetic energy at start = (1/2) m v^2

= (1/2)(3.5)(7.9)^2
= 109 Joules

At the maximum displacement, it stops, so that kinetic energy has all gone into stretching the spring which stores it as potential energy
Potential energy in spring = (1/2) k x^2
so
(1/2) ( 725) * x^2 = 109 Joules
x^2 = .301
x = .549 meters
= 55 cm

To find the maximum displacement of the object, we can use the concept of energy conservation.

The total energy of the system is conserved, and it is given by the sum of the kinetic energy and the potential energy:

E = KE + PE

Initially, when the object is at the equilibrium position, it has no potential energy but only kinetic energy. The equation becomes:

E = KE_initial + PE_initial

As the object moves upward, it gains potential energy at the expense of the kinetic energy. At the maximum displacement, the object momentarily stops, so its kinetic energy becomes zero while all of its initial kinetic energy is converted into potential energy:

E = KE_final + PE_max

Since the total energy E is conserved, we can equate the two expressions:

KE_initial + PE_initial = KE_final + PE_max

The initial kinetic energy is given by:

KE_initial = (1/2) * m * v^2

where m is the mass of the object and v is the initial velocity. Plugging in the values:

KE_initial = (1/2) * 3.5 kg * (7.9 m/s)^2

Next, let's find the potential energy at the maximum displacement. The potential energy of a spring is given by:

PE_max = (1/2) * k * x^2

where k is the spring constant and x is the displacement from the equilibrium position. We want to find x, so let's rearrange the equation:

x = sqrt((2 * PE_max) / k)

Plugging in the values:

x = sqrt((2 * PE_max) / 725 N/m)

Since all of the initial kinetic energy is converted into potential energy at the maximum displacement, we can equate the two expressions:

(1/2) * 3.5 kg * (7.9 m/s)^2 = (1/2) * 725 N/m * x^2

Simplifying the equation:

(3.5 kg * (7.9 m/s)^2) / (725 N/m) = x^2

x^2 = 3.5 kg * (7.9 m/s)^2 / 725 N/m

Taking the square root of both sides, we can find the maximum displacement x:

x = sqrt(3.5 kg * (7.9 m/s)^2 / 725 N/m)

Now we can calculate x:

x = sqrt(3.5 * 7.9^2 / 725) = 0.651 m = 65.1 cm

Therefore, the maximum displacement of the object is 65.1 cm.

To find the maximum displacement of the object, we need to analyze the energy of the system.

The total mechanical energy of a mass-spring system consists of two forms: kinetic energy (KE) and potential energy (PE).

The kinetic energy (KE) is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

The potential energy (PE) of an object in a spring is given by the equation PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.

In this case, the initial velocity of the object is 7.9 m/s, and the mass of the object is 3.5 kg. Therefore, the initial kinetic energy (KEi) can be calculated as:

KEi = (1/2) * 3.5 kg * (7.9 m/s)^2

To find the displacement at maximum height, we need to convert the initial kinetic energy to potential energy. Since energy is conserved in this system, the total mechanical energy is constant.

KEi = PE

(1/2) * 3.5 kg * (7.9 m/s)^2 = (1/2) * 725 N/m * x^2

Now, we can solve for x, the displacement at maximum height:

x^2 = [(3.5 kg * (7.9 m/s)^2) / (725 N/m)]
x^2 = 30.855

Taking the square root of both sides:

x ≈ 5.55 m

Finally, we convert the displacement from meters to centimeters:

x ≈ 5.55 m * 100 cm/m ≈ 555 cm

Therefore, the maximum displacement of the object is approximately 555 cm.