The mass m = 14 kg shown in the figure is displaced a distance x to the right from its equilibrium position. (Take k1 = 113 N/m and k2 = 198 N/m.)

(a) What is the net force acting on the mass? (Use any variable or symbol stated above along with the following as necessary: k_eff for the effective spring constant.)
Fnet =

What is the effective spring constant?


(b) What will the frequency of the oscillation be when the mass is released?
f =

(c) If x = 0.1 m, what is the total energy of the mass-spring system after the mass is released, and what is the maximum velocity of the mass?
Etotal =
vmax =

To find the net force acting on the mass, we need to find the force exerted by each spring and add them up.

The force exerted by a spring can be calculated using Hooke's Law, which states that the force is proportional to the displacement from the equilibrium position. The formula for the force exerted by a spring is:

F = -kx

Where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, we have two springs with different spring constants. To find the net force, we need to find the forces exerted by each spring and add them up.

The force exerted by the first spring (k1) is:

F1 = -k1x

The force exerted by the second spring (k2) is:

F2 = -k2x

To find the net force, we add the forces:

Fnet = F1 + F2 = -k1x - k2x

To simplify, we can factor out the x:

Fnet = -x(k1 + k2)

So the net force acting on the mass is -x times the sum of the spring constants (k1 + k2).

Next, let's find the effective spring constant.

The effective spring constant (k_eff) is a single spring constant that would produce the same net force as the two springs combined. It can be found using the formula:

1/k_eff = 1/k1 + 1/k2

Rearranging the formula:

k_eff = 1 / (1/k1 + 1/k2)

Let's calculate the effective spring constant using the given values:

k1 = 113 N/m
k2 = 198 N/m

k_eff = 1 / (1/113 + 1/198)
= 1 / (0.0088 + 0.0051)
= 1 / 0.0139
≈ 71.942 N/m

The effective spring constant is approximately 71.942 N/m.

Now let's move on to part (b) and find the frequency of oscillation when the mass is released.

The frequency of oscillation (f) can be calculated using the formula:

f = 1 / (2π) * √(k_eff / m)

Where f is the frequency, k_eff is the effective spring constant, and m is the mass.

Let's calculate the frequency using the given values:

m = 14 kg
k_eff ≈ 71.942 N/m

f = 1 / (2π) * √(71.942 / 14)
≈ 1 / (2π) * √5.13871
≈ 0.459 Hz

The frequency of the oscillation when the mass is released is approximately 0.459 Hz.

Finally, let's move on to part (c) and find the total energy of the mass-spring system after the mass is released and the maximum velocity of the mass.

The total energy of the system (E_total) can be calculated using the formula:

E_total = (1/2) k_eff x^2

Where E_total is the total energy, k_eff is the effective spring constant, and x is the displacement from the equilibrium position.

Let's calculate the total energy using the given displacement:

x = 0.1 m
k_eff ≈ 71.942 N/m

E_total = (1/2) * 71.942 * (0.1)^2
= 0.35971 J

The total energy of the mass-spring system after the mass is released is approximately 0.35971 J.

The maximum velocity of the mass (v_max) can be calculated using the formula:

v_max = √(k_eff / m) * x

Where v_max is the maximum velocity, k_eff is the effective spring constant, m is the mass, and x is the displacement from the equilibrium position.

Let's calculate the maximum velocity using the given values:

m = 14 kg
k_eff ≈ 71.942 N/m
x = 0.1 m

v_max = √(71.942 / 14) * 0.1
≈ √5.13871 * 0.1
≈ 0.716 m/s

The maximum velocity of the mass is approximately 0.716 m/s.