a mass m= 10kg is hanging from the end of a spring. intially the spring is unstretched when the mass is pushed perfectly downward with an initial velocity of 1 m/s. the spring constant is k=400n/m and the acceleration of gravity is g=9.81m/s^2. find the lowest position that the block reaches before it reverses direction and find the maximum speed that the mass attains along its descent
To find the lowest position that the block reaches before it reverses direction, we can use the conservation of mechanical energy. The mechanical energy of the system (mass hanging from the spring) is conserved if we neglect any energy losses due to friction or air resistance.
The mechanical energy is given by:
E = 1/2 k x^2 + mgh
Where:
k is the spring constant (400 N/m),
x is the displacement of the mass from its equilibrium position,
m is the mass (10 kg),
g is the acceleration due to gravity (9.81 m/s^2),
h is the height of the mass above the reference point.
Initially, when the mass is pushed perfectly downward, the spring is unstretched. This means x = 0. The mass starts with an initial velocity of 1 m/s, so its initial kinetic energy is:
KE_initial = 1/2 mv^2 = 1/2 (10 kg) (1 m/s)^2 = 5 J
At the lowest position, the block comes to a momentary stop and reverses its direction. This means its final kinetic energy is 0 J. Since there is no loss in mechanical energy, we can equate the initial kinetic energy to the potential energy at the lowest point:
KE_initial = PE_lowest
5 J = 1/2 k x^2 + mgh
Since x = 0, the first term (1/2 k x^2) becomes zero. We can simplify the equation:
5 J = mgh
Now, solve for h:
h = 5 J / (mg) = 5 J / (10 kg * 9.81 m/s^2) ≈ 0.051 m
Therefore, the lowest position that the block reaches before it reverses direction is approximately 0.051 meters above the reference point.
To find the maximum speed that the mass attains along its descent, we can use the conservation of mechanical energy again. At the highest point (when the mass is momentarily at rest), all of its mechanical energy is in the form of potential energy. At the lowest point, all of its mechanical energy is in the form of kinetic energy.
At the highest point, the potential energy is maximum and the kinetic energy is zero:
PE_highest = mgh = 10 kg * 9.81 m/s^2 * h
At the lowest point, the potential energy is zero and the kinetic energy is maximum:
KE_max = 1/2 mv_max^2
Since there is no loss in mechanical energy, we can equate the potential energy at the highest point to the kinetic energy at the lowest point:
PE_highest = KE_max
10 kg * 9.81 m/s^2 * h = 1/2 m v_max^2
Simplifying the equation:
9.81 m/s^2 * h = 1/2 v_max^2
Now, solve for v_max:
v_max = √(2 * 9.81 m/s^2 * h) = √(2 * 9.81 m/s^2 * 0.051 m) ≈ 1.43 m/s
Therefore, the maximum speed that the mass attains along its descent is approximately 1.43 meters per second.