A body of mass m is suspended from a spring with spring constant k and the spring is stretched 0.1. If two identical bodies of mass m/2 are suspended from a spring with the same spring constant k, how much will the spring stretch? Explain your answer.
The first case is when the mass is just pulling the spring downwards while the second case is the 2 m/2 masses connected to a pulley which pulls the spring horizontally.
Why isn't the answer 0.1m? Don't the forces still add up to the same as the first case?
A body of mass
In the first case, when a body of mass m is suspended from the spring and the spring is stretched by 0.1 units, the force applied by the mass stretches the spring based on Hooke's Law. Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.
The force applied by the mass is given by F = mg, where m is the mass and g is the acceleration due to gravity. Since the spring is stretched by 0.1 units, we can write the equation as F = k * x, where k is the spring constant and x is the displacement from the equilibrium position.
By substituting the values, we have mg = k * 0.1.
Now let's consider the second case where two identical masses of m/2 are suspended from the same spring. The force applied by each mass is still given by F = mg/2, as each mass is now m/2. Since both masses are identical, their individual displacements will be the same.
Now, to find out how much the spring stretches in this case, we need to find the total force applied to the spring. Since both masses are acting in the same direction (downwards), the total force will be the sum of individual forces, which is F_total = (mg/2) + (mg/2) = mg.
Using the same formula, F = k * x, we have mg = k * x2 (where x2 is the displacement in the second case).
Comparing the two cases, we can see that mg = k * 0.1 (first case) and mg = k * x2 (second case). Since mg is common to both equations, we can conclude that 0.1 = x2.
Therefore, the spring will stretch by 0.1 units in the second case as well. The answer is not 0.1m because the effective force acting on the spring is still based on the total mass (mg) rather than the individual masses (m/2). The combination of two masses balanced through a pulley does not change the effective force acting on the spring.