A single mass (m1 = 3.1 kg) hangs from a spring in a motionless elevator. The spring constant is k = 338 N/m.

1. What is the distance the spring is stretched from its unstretched length?
2. Now, three masses (m1 = 3.1 kg, m2 = 9.3 kg and m3 = 6.2) hang from three identical springs in a motionless elevator. The springs all have the same spring constant given above.

What is the magnitude of the force the bottom spring exerts on the lower mass?
3. What is the distance the middle spring is stretched from its equilibrium length?

2. F=ma=6.2(9.8)=60.76N

*If you are calculating with an additional acceleration, you will add that to g.

F=-kx=ma

m=m2+m3=9.3+6.2
a = g+elevator acceleration = -9.8
k=338
-9.8(15.5)=-338x
x=0.449m

1. Why did the spring go to therapy? Because it was feeling a little stretched out!

To calculate the distance the spring is stretched from its unstretched length, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The formula to calculate the displacement is given by x = (F / k), where x is the displacement, F is the force exerted by the mass, and k is the spring constant.

So, in this case, the force exerted by the mass is simply mg, where g is the acceleration due to gravity. Plugging in the values, x = (mg / k).

x = (3.1 kg)(9.8 m/s^2) / 338 N/m
x ≈ 0.089 m

Therefore, the spring is stretched by approximately 0.089 meters.

2. The bottom spring is really carrying the weight on its shoulders, isn't it?
To calculate the magnitude of the force the bottom spring exerts on the lower mass, we just need to consider the downward force acting on the mass. The downward force is simply the weight of the mass, which is given by F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity.

So, in this case, the magnitude of the force is F = (9.3 kg)(9.8 m/s^2).

F ≈ 91.14 N

Therefore, the bottom spring exerts a force of approximately 91.14 Newtons on the lower mass.

3. The middle spring is stretching its limits, just like all those pants in quarantine!
To calculate the distance the middle spring is stretched from its equilibrium length, we can again use Hooke's Law. We know that the force exerted by the middle spring should be equal to the sum of the forces exerted by the three masses. We can calculate the total force and then use the same formula as in question 1 to find the displacement.

The total force is simply the sum of the weights of the three masses, which is given by F_total = m1g + m2g + m3g.

So, in this case, the distance the middle spring is stretched is given by x = (F_total / k).

x = [(3.1 kg)(9.8 m/s^2) + (9.3 kg)(9.8 m/s^2) + (6.2 kg)(9.8 m/s^2)] / 338 N/m

x ≈ 0.640 m

Therefore, the middle spring is stretched by approximately 0.640 meters.

To solve these problems, 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. The equation for Hooke's Law is given by:

F = -kx

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

1. To find the distance the spring is stretched, we need to find the displacement (x). We can rearrange Hooke's Law to solve for x:

x = -F/k

The mass hanging from the spring experiences the force due to gravity, given by:

F = mg

Substituting this into the equation for x:

x = -(mg)/k

Plugging in the values given (m1 = 3.1 kg and k = 338 N/m):

x = - (3.1 kg)(9.8 m/s^2) / (338 N/m)

Solving this equation will give us the distance the spring is stretched.

2. In this case, we have three masses hanging from three identical springs. The force exerted by each spring is given by Hooke's Law:

F = -kx

We need to find the magnitude of the force exerted by the bottom spring on the lower mass (m3). The force exerted by the spring is equal in magnitude but opposite in direction to the force experienced by the lower mass. Therefore, we can use the equation for force due to gravity to calculate the magnitude of the force exerted by the bottom spring on the lower mass:

F = m3g

Substituting this into Hooke's Law:

-m3g = -kx

Rearranging the equation, we can solve for x:

x = (m3g)/k

By plugging in the values given (m3 = 6.2 kg and k = 338 N/m), we can find the distance the middle spring is stretched.

3. Similarly, to find the distance the middle spring is stretched, we can use Hooke's Law:

F = -kx

The force exerted by each spring is equal in magnitude but opposite in direction to the force experienced by the mass it is attached to. Therefore, we can use the equation for force due to gravity to calculate the magnitude of the force exerted by the middle spring on the middle mass (m2):

F = m2g

Substituting this into Hooke's Law:

-m2g = -kx

Rearranging the equation, we can solve for x:

x = (m2g)/k

By plugging in the values given (m2 = 9.3 kg and k = 338 N/m), we can find the distance the middle spring is stretched.

1. Given that known variable k = 338 N/m and F=mg=6.2(9.8), we will use Hooke's Law(F=-kx) to calculate the stretched distance for the spring. F=mg=-kx which is 60.76=338x. Therefore, x = 0.18m