A 800kg safe is 2.3m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 52cm.What is 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 is expressed as:
F = -k * x
Where:
F is the force exerted by the spring (in Newtons),
k is the spring constant (in Newtons per meter),
x is the displacement of the spring from its equilibrium position (in meters).
In this case, the safe falls and hits the spring, causing it to compress. We know the weight of the safe and the distance it falls (height above the spring).
First, let's calculate the gravitational potential energy (GPE) of the safe before it falls. We can use the formula:
GPE = m * g * h
Where:
m is the mass of the safe (in kilograms),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h is the height the safe is above the spring (in meters).
GPE = 800 kg * 9.8 m/s^2 * 2.3 m
GPE ≈ 17744 J
When the safe hits the spring, it transfers its potential energy into compressing the spring. The spring potential energy is given by:
SPE = (1/2) * k * x^2
Where:
k is the spring constant (in Newtons per meter),
x is the displacement of the spring (compression) (in meters).
In this case, the spring compresses by 52 cm, which is 0.52 meters.
SPE = (1/2) * k * (0.52 m)^2
Since the gravitational potential energy equals the spring potential energy:
GPE = SPE
17744 J = (1/2) * k * (0.52 m)^2
Now we can solve for k:
k = (2 * GPE) / x^2
k = (2 * 17744 J) / (0.52 m)^2
k ≈ 127345.15 N/m
Therefore, the spring constant of the spring is approximately 127345.15 N/m.