In a room that is 2.69 m high, a spring (unstrained length = 0.28 m) hangs from the ceiling. A board whose length is 2.22 m is attached to the free end of the spring. The board hangs straight down, so that its 2.22-m length is perpendicular to the floor. The weight of the board (145 N) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?
Elongation of the spring is
x=2.69 –(2.22+0.28) = 0.19 m
mg= kx
k=mg/x- 145/0.19=763.16 N/m
To calculate the spring constant (k) of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
First, let's determine the displacement of the spring. We know that the length of the board is 2.22 m, and the unstrained length of the spring is 0.28 m. So, the displacement of the spring is given by:
Displacement = Length of board - Unstrained length of spring
Displacement = 2.22 m - 0.28 m
Displacement = 1.94 m
Now, let's calculate the spring constant using Hooke's Law. According to Hooke's Law, the force (F) exerted by a spring can be calculated as:
F = k * x
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, the weight of the board (145 N) provides the force and the displacement of the spring is 1.94 m. So, we can rewrite Hooke's Law as:
145 N = k * 1.94 m
Now, solve for the spring constant (k):
k = 145 N / 1.94 m
k ≈ 74.74 N/m
Therefore, the spring constant of the spring is approximately 74.74 N/m.