A massless spring with spring constant 17.1 N/m hangs vertically. A body of mass 0.210 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. How far below the initial position does the body descend?
To find how far below the initial position the body descends, we can use the concept of potential energy in a spring.
The potential energy (PE) in a spring is given by the formula:
PE = 0.5 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring was initially unstretched, so the equilibrium position is the starting point of the body.
Since the spring is hanging vertically, the weight of the body is balanced by the spring force when the body is at equilibrium. We can use this information to determine the displacement.
The weight of the body is given by:
Weight = mass * acceleration due to gravity
Weight = 0.210 kg * 9.8 m/s^2 = 2.058 N
The spring force is equal to the weight of the body when it is at equilibrium. So, at the equilibrium position:
Spring Force = Weight
Using Hooke's Law, we have:
Spring Force = k * x
From the given information, we know that the spring constant (k) is 17.1 N/m.
Substituting the values, we have:
17.1 N/m * x = 2.058 N
Solving for x:
x = 2.058 N / 17.1 N/m = 0.120 m
Therefore, the body descends 0.120 m below the initial position when released.