two springs of constants 2.0*10^3 Nm and 3.0*10^3 Nm are connected in tandem, and a mass of 5.9kg hangs vertically from this spring. 1.) by what amount does the mass stretch the combined string? 2.) By what amount does the mass stretch each individual string?

x=mg/k

spring 1, 5.9*9.81/2000=?
spring 2, 5.9*9.81/3000=?

combined= spring 1 + spring 2

I answered my own question./

for springs connected in tandem

1/k=1/k1 +1/k2 =>
k=1200 N/m
the stretch is
x=mg/k=5.9•9.81/1200 =

To answer these questions, we can use Hooke's law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.

Let's start with the first question:

1. By what amount does the mass stretch the combined string?

To find the stretch of the combined springs, we need to find the effective spring constant (k_eff) of the system. When two springs are connected in tandem, their effective spring constant is given by:

1/k_eff = 1/k₁ + 1/k₂

where k₁ and k₂ are the spring constants of the individual springs.

Let's calculate k_eff:

k₁ = 2.0 × 10^3 N/m
k₂ = 3.0 × 10^3 N/m

1/k_eff = 1/k₁ + 1/k₂
1/k_eff = 1/(2.0 × 10^3 N/m) + 1/(3.0 × 10^3 N/m)
1/k_eff = (3.0 × 10^3 N/m + 2.0 × 10^3 N/m) / (2.0 × 10^3 N/m) × (3.0 × 10^3 N/m)
1/k_eff = 5.0 × 10^3 N/m / 6.0 × 10^3 N/m
1/k_eff = 5.0 / 6.0

Now, let's find k_eff:

k_eff = 6.0 × 10^3 N/m / 5.0
k_eff = 1.2 × 10^3 N/m

Now, we can use Hooke's law to find the amount the mass stretches the combined springs. Hooke's law is expressed as:

F = k_eff * x

where F is the force applied to the spring and x is the displacement.

The force acting on the spring is the weight of the mass:

F = m * g

where m is the mass (5.9 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

F = (5.9 kg) * (9.8 m/s^2)
F = 57.82 N

Now, we can find the displacement (x):

57.82 N = (1.2 × 10^3 N/m) * x

Solving for x:

x = 57.82 N / (1.2 × 10^3 N/m)
x ≈ 0.048 m

Therefore, the mass stretches the combined springs by approximately 0.048 meters.

Now, let's move on to the second question:

2. By what amount does the mass stretch each individual string?

To find the stretch of each individual spring, we can use Hooke's law with the respective spring constants.

For the first spring (k₁ = 2.0 × 10^3 N/m):

F = k₁ * x₁

where F is the force applied to the spring and x₁ is the displacement.

The force acting on the spring is the weight of the mass:

F = m * g

F = (5.9 kg) * (9.8 m/s^2)
F = 57.82 N

Now, we can find x₁:

57.82 N = (2.0 × 10^3 N/m) * x₁

Solving for x₁:

x₁ = 57.82 N / (2.0 × 10^3 N/m)
x₁ ≈ 0.029 m

Therefore, the mass stretches the first spring by approximately 0.029 meters.

For the second spring (k₂ = 3.0 × 10^3 N/m), we repeat the same calculations:

F = k₂ * x₂

where F is the force applied to the spring and x₂ is the displacement.

The force acting on the spring is the weight of the mass:

F = m * g

F = (5.9 kg) * (9.8 m/s^2)
F = 57.82 N

Now, we can find x₂:

57.82 N = (3.0 × 10^3 N/m) * x₂

Solving for x₂:

x₂ = 57.82 N / (3.0 × 10^3 N/m)
x₂ ≈ 0.019 m

Therefore, the mass stretches the second spring by approximately 0.019 meters.

To summarize:
1.) The mass stretches the combined springs by approximately 0.048 meters.
2.) The mass stretches the first spring by approximately 0.029 meters, and the second spring by approximately 0.019 meters.