A block (mass m) hangs from a spring (spring constant k). The block is released from rest a distance d above its equilibrium position.
(a) What is the speed of the block as it passes through the equilibrium point? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration due to gravity.)
v =
(b) What is the maximum distance below the equilibrium point that the block will reach? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration due to gravity.)
To find the speed of the block as it passes through the equilibrium point, we can use the principle of conservation of mechanical energy. At its highest point, all of the potential energy from being raised to a height d is converted into kinetic energy as it passes through the equilibrium point.
(a) Speed of the block as it passes through the equilibrium point:
We can find the potential energy at the starting position and the kinetic energy at the equilibrium point, equate them and solve for the speed.
Potential Energy at starting position = mass * gravity * height
Potential Energy = m * g * d
Kinetic Energy at equilibrium point = 1/2 * mass * velocity^2, where velocity is the speed
At the equilibrium point, the potential energy is completely converted into kinetic energy. So we can set the potential energy equal to the kinetic energy and solve for the speed.
m * g * d = 1/2 * m * v^2
Simplifying the equation, we get:
v^2 = 2 * g * d
Taking the square root of both sides, we find:
v = √(2 * g * d)
So the speed of the block as it passes through the equilibrium point is given by √(2 * g * d).
(b) Maximum distance below the equilibrium point that the block will reach:
At the highest point, where the block is released, all of the energy is in the form of potential energy. As the block descends, it converts potential energy into kinetic energy. When it reaches the maximum distance below the equilibrium point, all of the potential energy is converted into kinetic energy.
To find the maximum distance below the equilibrium point, we can equate the potential energy at the starting position to the kinetic energy at the maximum distance.
Potential Energy at starting position = mass * gravity * height
Potential Energy = m * g * d
Kinetic Energy at maximum distance = 1/2 * mass * velocity^2, where velocity is the speed
Setting the potential energy equal to the kinetic energy, we have:
m * g * d = 1/2 * m * v^2
Now, we can solve for the maximum distance by rearranging the equation:
d = v^2 / (2 * g)
So the maximum distance below the equilibrium point that the block will reach is given by v^2 / (2 * g).