Question

a mass of m= 5.00 kg box is attatched to a spring that is fixed to a wall. the spring constant k=250 N/m. the spring is fully compressed to position I equal to 10 cm from its equilibrium position II and then let go.

a) what are the kinetic and potential energies of the mass with fuly compressed spring at position I?

b) what is the potential energy of the compressed spring as it is 8 cm from equilibrium?

C) what is the speed of the object when it is 8 cm from equilibrium?

d) what is the speed of the object in equilibrium?

Answer 1

Answer 2

a) The potential energy of the compressed spring can be calculated as:

U = (1/2) kx^2

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

At position I, x = 0.1 m, so the potential energy is:

U = (1/2) (250 N/m) (0.1 m)^2 = 1.25 J

Since the mass is at rest at position I, its kinetic energy is zero.

b) When the spring is 8 cm from equilibrium (position III), the displacement is x = 0.08 m. The potential energy of the compressed spring can be calculated as:

U = (1/2) kx^2 = (1/2) (250 N/m) (0.08 m)^2 = 0.8 J

c) At position III, the mass has potential energy from the compressed spring, but also kinetic energy since it is in motion. The total mechanical energy of the system (the mass and the spring) is conserved, so:

E = U + K

where E is the total mechanical energy, U is the potential energy, and K is the kinetic energy.

Since the potential energy at position III is 0.8 J and there is no kinetic energy at position III, the total mechanical energy at position III is:

E = 0.8 J + 0 J = 0.8 J

At position III, all of the potential energy from the compressed spring has been converted to kinetic energy. Therefore, the kinetic energy at position III is:

K = E - U = 0.8 J - 0.8 J = 0 J

Using the conservation of energy, we can set the initial total mechanical energy (at position I) equal to the total mechanical energy at position III:

E_i = E_f

(1/2) kx_i^2 = (1/2) mv_f^2

where x_i is the initial displacement (0.1 m), m is the mass (5.00 kg), v_f is the speed at position III (which we want to find), and k is the spring constant (250 N/m).

Solving for v_f, we get:

v_f = sqrt((k/m) * x_i^2) = sqrt((250 N/m) / 5.00 kg * (0.1 m)^2) = 1.58 m/s

d) At equilibrium (position II), the displacement is x = 0 m, so the potential energy of the spring is zero. Using the conservation of energy, we can set the initial total mechanical energy (at position I) equal to the total mechanical energy at equilibrium:

E_i = E_eq

(1/2) kx_i^2 = (1/2) mv_eq^2

where x_i is the initial displacement (0.1 m), m is the mass (5.00 kg), v_eq is the speed at equilibrium (which we want to find), and k is the spring constant (250 N/m).

Solving for v_eq, we get:

v_eq = sqrt((k/m) * x_i^2) = sqrt((250 N/m) / 5.00 kg * (0.1 m)^2) = 1.58 m/s

Therefore, the speed of the object in equilibrium is also 1.58 m/s.