an object is attached to the lower end of a 100-coil spring hanging from the ceiling.the spring stretches by 0.160m. the spring is then cut into two identical springs of 50 coils each. as the drawing shows, each spring is attached between the cieling and the object by how much does each spring stretch?
in preparation for shooting a pinball machine,a spring (k=675n/m) is compressed
To find out how much each spring stretches when it is cut into two identical springs with 50 coils each, we need to consider the principle of conservation of energy.
The potential energy stored in a spring is given by the formula:
PE = (1/2) * k * x^2
Where PE is the potential energy, k is the spring constant, and x is the displacement or stretch of the spring from its equilibrium position.
Now, let's analyze the situation step by step:
1. Initially, we have a 100-coil spring hanging from the ceiling, and it stretches by 0.160m. Let's denote this initial displacement as x1.
2. Since the spring is cut into two identical springs, each with 50 coils, the spring constant (k) remains the same for each spring.
3. We need to find the new displacement for each spring. Let's denote it as x2.
According to the conservation of energy, the potential energy before cutting the spring must be equal to the potential energy after cutting it into two identical springs.
So, we can set up an equation:
(1/2) * k * x1^2 = 2 * (1/2) * k * x2^2
The factor of 2 on the right-hand side accounts for the fact that we now have two identical springs.
Now, let's simplify the equation:
k * x1^2 = k * x2^2
Divide both sides of the equation by k:
x1^2 = x2^2
Take the square root of both sides:
x1 = x2
Therefore, the displacement or stretch of each spring after cutting will be the same as the initial displacement or stretch of the original 100-coil spring, which is 0.160m.