A 1100 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 40 cm. What is the spring constant of the spring?

I don't understand what to do here

We know that potential energy is m*g*h (h = 2.0 + .4 = 2.4m)

and Hooke's formula is 1/2*k*x^2

In this problem when the safe drops its potential energy is turned into kinetic, therefore m*g*h = 1/2*k*x^2

Solve for k:
(2*m*g*h) / x^2

Well, it sounds like you have quite the "springy" situation on your hands! To calculate the spring constant, we can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the amount it is compressed.

In this case, we know that the safe weighs 1100 kg, so the force exerted by gravity on the safe is given by F = m * g, where m is the mass of the safe and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, as the safe falls and compresses the spring, the potential energy from gravity is converted into elastic potential energy stored in the spring. Using the equation for potential energy stored in a spring, PE = (1/2) * k * x^2, where k is the spring constant and x is the distance the spring is compressed.

We can equate the gravitational potential energy (m * g * h) to the elastic potential energy to get:

m * g * h = (1/2) * k * x^2

Plugging in the known values, we have:

(1100 kg) * (9.8 m/s^2) * (2.0 m) = (1/2) * k * (0.4 m)^2

Now we just solve for k:

(k) = (1100 kg * 9.8 m/s^2 * 2.0 m) / (0.5 * (0.4 m)^2)

Calculating this out will give you the spring constant k for the heavy-duty spring!

To find the spring constant of the spring, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Hooke's Law can be written as: F = -kx

Where:
F is the force exerted by the spring,
k is the spring constant,
x is the displacement from the equilibrium position.

In this scenario, the safe falls and compresses the spring. We know the safe's mass (1100 kg), the height it falls from (2.0 m), and the displacement of the spring when compressed (40 cm or 0.40 m).

To determine the force exerted on the spring, we need to calculate the gravitational potential energy of the safe when it was at its initial height.

The potential energy (PE) of an object is given by the formula: PE = mgh

Where:
m is the mass of the object,
g is the acceleration due to gravity (which is approximately 9.8 m/s^2),
h is the height of the object.

Calculating the potential energy of the safe, we have:
PE = (1100 kg)(9.8 m/s^2)(2.0 m)
PE = 21,560 J

Since the spring absorbed all the potential energy when compressed, this potential energy is equal to the elastic potential energy stored in the spring.

Elastic potential energy (PE_elastic) is given by the formula: PE_elastic = (1/2)kx^2

Rearranging this equation, we get:
k = (2PE_elastic) / x^2

Substituting the known values:
k = (2 * 21,560 J) / (0.40 m)^2
k = 2,156,000 N/m

So, the spring constant (k) of the spring is approximately 2,156,000 N/m.

mgh = 1/2 kx^2

solve for k

1.1(9.8)(2.0)

=21.56