A light spring with spring constant 1 200 N/m is hung from

an elevated support. From its lower end hangs a second
light spring, which has spring constant 1 800 N/m. An
object of mass 1.50 kg is hung at rest from the lower end of
the second spring. (a) Find the total extension distance of
the pair of springs. (b) Find the effective spring constant
of the pair of springs as a system. We describe these springs
as in series.

Springs in series add like resistors in parallel.

1/Keff = 1/K1 + 1/K2
= 1/1200 + 1/1800
= 5/3600 = 1/720

Keff = 720 N/m
Is the effective spring constant of the pair of springs in series.

Extension = M*g/Keff

To find the total extension distance of the pair of springs, we need to consider the forces acting on the object.

(a) The force exerted by the first spring (F1) can be calculated using Hooke's Law:

F1 = k1 * x1

where k1 is the spring constant of the first spring and x1 is the extension distance of the first spring.

The force exerted by the second spring (F2) can also be calculated using Hooke's Law:

F2 = k2 * (x1 + x2)

where k2 is the spring constant of the second spring and x2 is the extension distance of the second spring.

Since the object is in equilibrium, the total force acting on it must be zero:

F1 + F2 = 0

Plugging in the values and re-arranging the equation:

k1 * x1 + k2 * (x1 + x2) = 0

Now, let's substitute the given values:

1200 * x1 + 1800 * (x1 + x2) = 0 (Eq.1)

(b) The effective spring constant of the pair of springs can be calculated using the equation:

1/keff = 1 / k1 + 1 / k2

Substituting the given values:

1/keff = 1/1200 + 1/1800

To find keff, we take the reciprocal of both sides:

keff = 1 / (1/1200 + 1/1800)

Now, let's solve for the values of x1, x2, and keff.

To solve Eq.1, we can simplify it:

1200 * x1 + 1800 * x1 + 1800 * x2 = 0
3000 * x1 + 1800 * x2 = 0
500 * x1 + 300 * x2 = 0
Divide both sides of the equation by 100 to simplify it further:

5 * x1 + 3 * x2 = 0 (Eq.2)

Now, we can solve Eq.2 simultaneously with Eq.1:

5 * x1 + 3 * x2 = 0 (Eq.2)
1200 * x1 + 1800 * (x1 + x2) = 0 (Eq.1)

To do this, we can use the method of substitution:

Solve Eq.2 for x1:
x1 = -3/5 * x2

Substitute this value of x1 into Eq.1:

1200 * (-3/5 * x2) + 1800 * (x1 + x2) = 0

Simplify the equation:

-720 * x2 + 1800 * (-3/5 * x2 + x2) = 0
-720 * x2 + 1800 * (-3/5 + 2/5) * x2 = 0
-720 * x2 + 1800 * (-1/5) * x2 = 0
-720 * x2 - 360 * x2 = 0
-1080 * x2 = 0

Divide both sides of the equation by -1080 to solve for x2:

x2 = 0

Now, substitute the value of x2 back into Eq.2 to solve for x1:

5 * x1 + 3 * 0 = 0
5 * x1 = 0

Therefore, x1 = 0 and x2 = 0, which means there is no extension in either of the springs, and the object is at rest.

For the effective spring constant, keff:

keff = 1 / (1/1200 + 1/1800)
keff = 1 / ((3/3600) + (2/3600))
keff = 1 / (5/3600)
keff = 3600/5
keff = 720 N/m

To find the total extension distance of the pair of springs, we need to sum up the individual extensions of each spring.

Let's assume that the extension of the first spring is x1 and the extension of the second spring is x2.

The force on the first spring can be calculated using Hooke's Law: F1 = k1 * x1, where k1 is the spring constant of the first spring (1,200 N/m).

The force on the second spring can be calculated similarly: F2 = k2 * x2, where k2 is the spring constant of the second spring (1,800 N/m).

The total force on the system is equal to the weight of the object hanging at rest, which is given by: F_total = m * g, where m is the mass of the object (1.50 kg) and g is the acceleration due to gravity (9.8 m/s^2).

Since the springs are in series, the total force is the sum of the forces on each spring: F_total = F1 + F2.

Substituting the expressions for F1 and F2, we get: k1 * x1 + k2 * x2 = m * g.

Now we can solve this equation for the total extension distance x1 + x2.

(a) Find the total extension distance:
Rearranging the equation, we have: x1 + x2 = (m * g) / (k1 + k2). Plugging in the given values, we get:
x1 + x2 = (1.50 kg * 9.8 m/s^2) / (1,200 N/m + 1,800 N/m).
Calculating the numerator and denominator separately, we find:
x1 + x2 = (14.7 kg*m/s^2) / (3,000 N/m).
Finally, dividing the numerator by the denominator, we obtain the total extension distance:
x1 + x2 = 0.0049 m.

(b) Find the effective spring constant:
The effective spring constant of a system of springs in series can be calculated by:
1/k_effective = 1/k1 + 1/k2.
Substituting the values, we get:
1/k_effective = 1/1,200 N/m + 1/1,800 N/m.
Simplifying this equation, we find:
1/k_effective = (1,800 N/m + 1,200 N/m) / (1,200 N/m * 1,800 N/m).
Calculating the numerator and denominator separately, we get:
1/k_effective = 3,000 N/m / (2,160,000 N^2/m^2).
Finally, we take the reciprocal of both sides to find the effective spring constant:
k_effective = (2,160,000 N^2/m^2) / 3,000 N/m.
Simplifying this expression, we find:
k_effective = 720 N/m.