Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series, as shown in (Figure 1) .

Part A: When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x?

k1
I
k2
I
m
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x

To calculate the amount of stretch, x, we need to use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.

First, let's calculate the effective spring constant, keff, for the springs connected in series.
When springs are connected in series, the effective spring constant is given by:

1/keff = 1/k1 + 1/k2

Substituting the given values, we have:

1/keff = 1/175 + 1/270

Calculating this expression gives us:
1/keff ≈ 0.0057

Taking the reciprocal of both sides, we find:
keff ≈ 174.8 N/m

Next, we can calculate the force exerted by the combined springs using Hooke's Law. The equation for Hooke's Law is:

F = keff * x

where F is the force exerted by the springs and x is the displacement or stretch.

Rearranging the equation to solve for x, we have:

x = F / keff

The force F can be calculated using Newton's second law. The weight of the mass is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2).

F = m * g

Substituting the given values, we have:

F = 0.50 kg * 9.8 m/s^2
F ≈ 4.9 N

Now we can calculate the amount of stretch, x:

x = 4.9 N / 174.8 N/m
x ≈ 0.028 m or 2.8 cm

Therefore, when a mass of 0.50 kg is attached to the springs connected in series, the amount of stretch, x, is approximately 0.028 m or 2.8 cm.

To find the amount of stretch, x, we can apply Hooke's Law for springs connected in series.

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring from its equilibrium position.

In this case, the force exerted by the two springs in series is equal to the force applied by the mass.

So, the equation we can use is:

F = k1 * x1 + k2 * x2

Where:
F is the force applied by the mass,
k1 and k2 are the force constants of the springs,
x1 and x2 are the displacements or stretches of the springs.

Let's substitute the given values into the equation:

F = (175 N/m) * x1 + (270 N/m) * x2

Since the two springs are connected in series, the amount of stretch, x, is the same for both springs. So, we can rewrite the equation as:

F = (175 N/m + 270 N/m) * x

Simplifying further:

F = 445 N/m * x

Now, let's substitute the mass value:

F = 445 N/m * x

0.50 kg * 9.8 m/s^2 = 445 N/m * x

4.9 N = 445 N/m * x

To find the value of x, we can rearrange the equation:

x = (4.9 N) / (445 N/m)

x ≈ 0.011(m) or 11.0 (mm)

Therefore, the amount of stretch, x, is approximately 0.011 m or 11.0 mm when a mass of 0.50 kg is attached to the springs.

the force is the same in each so add the extensions:

mg/k1 + mg/k2