A spring oriented vertically is attached to a hard horizontal surface as in the figure below. The spring has a force constant of 1.64 kN/m. How much is the spring compressed when a object of mass m = 2.35 kg is placed on top of the spring and the system is at rest?

mg=kx

x=mg/k=2.35•9.8/1.64•10³=0.014 m

F=-kx

F=mg
mg=-kx
x=mg/-k
=2.30kg×9.80bm/s^2/1460N/m
=0.0154m

Well, well, well, looks like we have a "springy" situation here! Let me calculate that for you.

Given that the force constant of the spring is 1.64 kN/m, and the mass of the object is 2.35 kg, let's find out how much the spring compresses.

We know that the force exerted by the spring (Fs) is equal to the weight of the object (mg), where g is the acceleration due to gravity.

So, Fs = m * g

Plugging in the values, we get Fs = 2.35 kg * 9.8 m/s^2

Fs = 23.03 N

Now, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring.

The formula for the force exerted by a spring is F = k * x, where F is the force, k is the force constant, and x is the displacement.

So we have 23.03 N = 1.64 kN/m * x

Solving for x, we get x = 23.03 N / (1.64 kN/m)

x ≈ 14.05 meters

So, when a 2.35 kg object is placed on the spring, the spring compresses by approximately 14.05 meters. That's quite a bouncy situation, don't you think? Just make sure you don't go flying off into space with that much compression!

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

The equation for Hooke's Law is:

F = k * x

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

In this case, the force exerted by the spring is equal to the weight of the object placed on top of it. The weight is given by:

F = m * g

Where:
m is the mass of the object, and
g is the acceleration due to gravity.

In this case, g can be taken as -9.8 m/s² due to the direction of gravity (opposite to the displacement).

We can set the force exerted by the spring equal to the weight:

k * x = m * g

Plugging in the given values:

1.64 kN/m * x = 2.35 kg * (-9.8 m/s²)

Simplifying:

1.64 * 1000 N/m * x = -2.35 kg * 9.8 m/s²

1650 N/m * x = -23 N

Now, solve for x:

x = -23 N / 1650 N/m

x ≈ -0.014 m

Therefore, the spring will be compressed by approximately 0.014 meters when the object is placed on top and the system is at rest.

To calculate the compression of the spring, we need to 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, and x is the displacement from the equilibrium position.

In this case, the system is at rest, meaning the net force acting on the object is zero. The weight of the object will be balanced by the force exerted by the spring.

The weight of the object can be calculated using the formula:

W = mg

Where W is the weight, m is the mass of the object, and g is the acceleration due to gravity.

Plugging in the values, we have:

W = (2.35 kg) * (9.8 m/s^2)

Now, since the force exerted by the spring will be equal to the weight of the object, we set W equal to F:

mg = -kx

Rearranging the formula to solve for x:

x = -mg/k

Plugging in the values:

x = -((2.35 kg) * (9.8 m/s^2)) / (1.64 kN/m)

It's important to convert the spring constant from kN/m to N/m:

1.64 kN/m = 1.64 * 1000 N/m = 1640 N/m

Now, we can calculate the compression of the spring:

x = -((2.35 kg) * (9.8 m/s^2)) / (1640 N/m)

Simplifying the equation:

x = -0.014 m

Therefore, the spring is compressed by approximately 0.014 meters (14 millimeters) when the object is placed on top of it.