A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.93m

Question

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.93m and whose unstreched length is 0.53m. Next the mass is pulled own to where the spring has a length of 1.05m given an initial speed upwards of 1.6. What is the maximum length of the spring during the motion that follows?

Answer 1

To find the maximum length of the spring during the motion that follows, we can use the principles of mechanical energy conservation.

The total mechanical energy of the system is conserved when no external forces (like friction or air resistance) are acting on it. In this case, the mechanical energy of the mass-spring system will be conserved.

The mechanical energy (E) of the system is the sum of kinetic energy (K) and potential energy (U):

E = K+U

The kinetic energy is given by:

K = 1/2mv²

where m is the mass of the object (0.4 kg) and v is the velocity of the object.

The potential energy in a spring-mass system is given by:

U = 1/2kx²

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

The difference between the initial and final mechanical energies is zero since energy is conserved:

E_initial​ − E_final​ = 0

Initially, the mass is at rest, so there's no kinetic energy (K_initial​ = 0).

The potential energy is given by the spring's displacement from its unstretched length (x_initial​ = 0.93m − 0.53m).

The final potential energy is zero when the spring is at its maximum elongation (U_final​ = 0).

At this point, the kinetic energy is given by the initial speed (v_final​ = 1.6m/s).

The equation becomes:

0 − [1/2k(0.93m−0.53m)²] = 1/2mv_final^2​

Now, we can solve for the spring constant, k:

1/2k(0.4m)² = 1/2(0.4 kg)(1.6m/s)²

Now, calculate the value of k:

k = (0.4 kg) (1.6 m/s)² / (0.4m)² ≈ 20 N/m

We have the spring constant, so now we can find the maximum elongation (x) of the spring during the motion.

At the maximum elongation, the kinetic energy is zero (K_final​=0), and all the mechanical energy is potential energy:

E_final​ = U_final​ = 1/2​kx²

Substitute the values:

0 = 1/2(20 N/m)x²

Solve for x:

x = √0/20N/m​ ​= 0 m

Therefore, the maximum length of the spring during the motion is the unstretched length, which is 0.53 m.