A mass, m1 = 5.5 kg, is in equilibrium while
connected to a light spring of constant k =
132 N/m that is fastened to a wall (see a). A
second mass, m2 = 10.9 kg, is slowly pushed
up against mass m1, compressing the spring
by the amount Ai = 0.17 m (see b). The
system is then released, causing both masses
to start moving to the right on the frictionless
surface. When m1 is at the equilibrium point,
m2 loses contact with m1 (see c) and moves
to the right with speed vmax.
(a)
(b)
(c)
(d)
Determine the value of vmax.
Answer in units of m/s
To determine the value of vmax, we can use the principle of conservation of mechanical energy. When the system is released, the potential energy stored in the compressed spring is converted into the kinetic energy of both masses. At the equilibrium point, all the potential energy has been converted to kinetic energy.
First, let's calculate the potential energy stored in the spring when it is compressed by Ai = 0.17 m. The potential energy stored in a spring is given by the equation:
PE = (1/2)kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
Substituting the values, we have:
PE = (1/2)(132 N/m)(0.17 m)^2
≈ 1.1236 J
Next, we can calculate the kinetic energy of the system at the equilibrium position. The kinetic energy is given by the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
Since the system is in equilibrium, the masses are connected, and they both have the same velocity. Let's denote this velocity as v_eq.
For mass m1 at the equilibrium point:
KE_1 = (1/2)m1v_eq^2
For mass m2 at the equilibrium point:
KE_2 = (1/2)m2v_eq^2
Since the sum of the kinetic energies must equal the potential energy stored in the spring, we have:
KE_1 + KE_2 = PE
Substituting the values, we have:
(1/2)m1v_eq^2 + (1/2)m2v_eq^2 = 1.1236 J
Simplifying the equation, we get:
(1/2)(m1 + m2)v_eq^2 = 1.1236 J
Now, we can solve for v_eq:
v_eq^2 = (2 * 1.1236 J) / (m1 + m2)
v_eq^2 = 2.2472 J / (5.5 kg + 10.9 kg)
v_eq^2 ≈ 0.0949 J/kg
Taking the square root of both sides, we get:
v_eq ≈ sqrt(0.0949 J/kg)
v_eq ≈ 0.3084 m/s
Since both masses lose contact at the equilibrium point, the velocity vmax of mass m2 is equal to v_eq.
Therefore, the value of vmax is approximately 0.3084 m/s.