An apple (m = 125g) is attached to a vertical spring (from a ceiling) that has a k value of 55 N/m. The apple is held up so the spring is un-stretched. When the apple is released, a) how far will the spring stretch? b) How fast will it be going when it has fallen 4.0cm?
To solve this problem, we need to use equations related to the behavior of springs and motion. Let's break it down into two parts: a) finding the stretch of the spring when the apple is released, and b) finding the velocity of the apple when it has fallen 4.0cm.
a) Finding the stretch of the spring:
The force exerted by a spring can be calculated using Hooke's Law, which states that the force (F) applied by a spring is equal to the spring constant (k) multiplied by the stretch (x) of the spring from its equilibrium position.
F = kx
Since the apple is in equilibrium before it's released, the force applied by the spring will be equal to the weight force acting on the apple.
Weight force (Fw) = mg
To find the stretch of the spring, we equate these two forces and solve for x:
mg = kx
Given:
Mass of the apple, m = 125g = 0.125kg
Spring constant, k = 55 N/m
Substituting the known values:
0.125kg * 9.8m/s^2 = 55N/m * x
Solving for x:
x = (0.125kg * 9.8m/s^2) / (55N/m)
x ≈ 0.0227m or 2.27cm
Therefore, the spring will stretch approximately 2.27cm.
b) Finding the velocity of the apple when it has fallen 4.0cm:
We can use the principles of conservation of energy to find the velocity of the apple when it has fallen a certain height.
The potential energy stored in the spring at its maximum stretch is given by:
Potential energy (PE) = 0.5kx^2
Since the potential energy will be converted entirely into kinetic energy as the apple falls, we can equate the potential energy to the kinetic energy to find the velocity.
Potential energy (PE) = Kinetic energy (KE)
0.5kx^2 = 0.5mv^2
Given:
x = 4.0cm = 0.04m
m = 0.125kg
k = 55 N/m
Substituting the known values:
0.5 * 55 N/m * (0.04m)^2 = 0.5 * 0.125kg * v^2
Solving for v:
v^2 = (55 N/m * (0.04m)^2) / 0.125kg
v^2 ≈ 0.1408 m^2/s^2
Taking the square root:
v ≈ 0.3758 m/s
Therefore, the velocity of the apple when it has fallen 4.0cm is approximately 0.3758 m/s.