a 0.1 kg mass is suspended at rest from a spring near the Earth's surface, the distance that the spring is stretched is measured to be 1.0 cm. What is the spring constant of the spring (remember the MKS units)?

A mass of 2 kg is attached to a spring with constant 18 N/m. How far is the spring stretched?

A spring with a spring constant of 28.9 N/m is attached to different masses, and is stretched 5cm. What is the force exerted on the spring?

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed.

1. In the first scenario, we have a 0.1 kg mass and the spring is stretched by 1.0 cm (or 0.01 m).

We can use the formula: F = k * x, where F is the force, k is the spring constant, and x is the distance stretched.

Since the mass is suspended at rest, the force acting on the mass is equal to its weight, which is given by the formula: F = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).

Setting the two equations equal and plugging in the values, we have: k * 0.01 = 0.1 * 9.8

To solve for k, we rearrange the equation: k = (0.1 * 9.8) / 0.01

Calculating this expression, we find that the spring constant is k = 98 N/m.

2. In the second scenario, we have a mass of 2 kg and a spring constant of 18 N/m.

Again, we can use the formula: F = k * x, where F is the force, k is the spring constant, and x is the distance stretched.

We are given the spring constant and need to find the distance stretched. Rearranging the formula, we have: x = F / k.

Plugging in the values, we have: x = 2 / 18.

Evaluating this expression, we find that the spring is stretched by approximately 0.11 m or 11 cm.

3. In the third scenario, we have a spring constant of 28.9 N/m and the spring is stretched by 5 cm (or 0.05 m).

Using the same formula: F = k * x, we can plug in the values to find the force exerted on the spring: F = 28.9 * 0.05.

Evaluating this expression, we find that the force exerted on the spring is approximately 1.45 N.

To calculate the spring constant (k) in the first question, we can use Hooke's Law: 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.

Given:
Mass (m) = 0.1 kg
Displacement (x) = 1.0 cm = 0.01 m

Now, let's solve for the spring constant (k):
F = -kx
F = -mass x acceleration due to gravity
kx = mg
k = (mg) / x
k = (0.1 kg x 9.8 m/s^2) / 0.01 m
k = 98 N/m

Therefore, the spring constant of the spring in the first question is 98 N/m.

To answer the second question, we can use the same equation: F = -kx. Rearranging the equation, we get:
x = -F / k.

Given:
Mass (m) = 2 kg
Spring constant (k) = 18 N/m

Let's solve for the displacement (x):
x = -F / k
x = -(mg) / k
x = -(2 kg x 9.8 m/s^2) / 18 N/m
x ≈ -1.078 m

The negative sign indicates that the displacement is in the opposite direction of the applied force. Therefore, the spring is stretched approximately 1.078 meters.

To answer the third question, we can use Hooke's Law: F = -kx.

Given:
Spring constant (k) = 28.9 N/m
Displacement (x) = 5 cm = 0.05 m

Let's calculate the force (F) exerted on the spring:
F = -kx
F = -(k)(0.05 m)
F = -(28.9 N/m)(0.05 m)
F = -1.45 N

Therefore, the force exerted on the spring in the third question is approximately 1.45 N.

Part 1: A M=0.1 kg mass weighs F = Mg = 0.98 Newtons. The spring constant is

k = F/X . If you put X in meters (0.01 m), do the computation and you will get k in MKS units of N/m

Part 2:
Use the same Hooke's law formula,
k = F/X
Rearrange it to get X= F/k
The do the calculation. They told you what k is. You will have to convert the 2 kh mass to weight force.

Part 3: F = k X
In this case they tell you that X = 0.05 m

Do the numbers.