A block is suspended from the end of a spring. An external force holds the block so that initially the spring is not stretched or compressed. When the block is released, it oscillates up and down between positions A and B. If the mass of the block is 0.5 kg and the spring constant is 10 N/m, how far will the block fall when it is released?
It will fall twice the equilibrium stretch distance,
2*M*g/k = 0.98 m, and oscillate about the equilibrium level, M*g/k, eventually settling there.
To find how far the block will fall when it is released, we need to determine the amplitude of its oscillation.
In this case, the block is initially at rest, so its total mechanical energy is purely potential energy. This potential energy is stored in the spring due to its displacement when the block is released.
The potential energy stored in a spring is given by the formula:
Potential energy (PE) = (1/2) * k * x^2
Where:
k is the spring constant (10 N/m in this case)
x is the displacement of the block from its equilibrium position (which is at the middle between positions A and B)
To find the displacement (x), we need to consider the forces acting on the block at positions A and B. At positions A and B, the block is at its maximum potential energy, and therefore at its maximum displacement.
At position A, the block has no kinetic energy, so its total mechanical energy is purely potential energy given by PE = (1/2) * k * x^2.
At position B, the block temporarily stops before changing direction and moving back towards position A. Again, its total mechanical energy is purely potential energy given by PE = (1/2) * k * x^2.
Setting the two equations equal to each other, we can solve for x:
(1/2) * k * x^2 = (1/2) * k * x^2
This equation shows that the potential energy at position A is equal to the potential energy at position B, so the displacement (x) is the same for both positions.
Now, we know that at position A, the potential energy is at its maximum. When the block is released, the potential energy is completely converted into kinetic energy as it falls towards its equilibrium position.
So, to find the maximum displacement (x), we can set the potential energy at position A equal to the kinetic energy at position B:
(1/2) * k * x^2 = (1/2) * m * v^2
Where:
m is the mass of the block (0.5 kg in this case)
v is the velocity of the block at position B (which we will solve for)
Since the block is at rest at position A, the velocity at position B, when the block is at its maximum displacement, is zero. Therefore, we can simplify the equation to:
(1/2) * k * x^2 = (1/2) * m * 0^2
Simplifying further, we get:
(1/2) * k * x^2 = 0
Since the left side of the equation can never be zero (given that k and x^2 are positive), the only solution is x = 0.
Therefore, the distance the block will fall when released is zero. It will start oscillating between positions A and B, with no net displacement from its initial position.