One of their side jobs is a safe delivery and installation. A pulley system is used to lift the safe 2 meters above the ground. But the rope snaps! To avoid cracking from any drops, a large, stiff spring is placed below the safe while it's lifted. If the 1200 kilograms safe hits the spring and compresses 17 centimeters, what's the spring constant of the spring?
To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be written as:
F = kx
Where:
F is the force applied on the spring
k is the spring constant
x is the displacement of the spring from its equilibrium position
In this scenario, the safe hits the spring, causing it to compress. The mass of the safe, m, is given as 1200 kilograms, and the displacement of the spring, x, is given as 17 centimeters.
First, we need to find the force exerted by the safe on the spring. We can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration:
F = m * g
Where:
m is the mass of the safe
g is the acceleration due to gravity
Since the safe is at rest, the force exerted by the safe on the spring is equal to the weight of the safe, which can be calculated as:
F = m * g
Substituting the values, we get:
F = 1200 kg * 9.8 m/s^2
Solving this equation, we find that the force exerted by the safe on the spring is 11760 Newtons.
Now, we can use Hooke's Law to find the spring constant, k. Rearranging the equation, we have:
k = F / x
Substituting the values, we get:
k = 11760 N / 0.17 m
Calculating this, we find that the spring constant of the spring is approximately 69,176.47 N/m.